NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Chapter 8 "Introduction to Trigonometry" contains four exercises. Below is a brief explanation of each exercise:

Exercise 8.1 - In this exercise, students are introduced to the basic trigonometric ratios - sine, cosine, and tangent, and their reciprocal functions. They learn how to find the values of these ratios for acute angles in a right triangle. The exercise also covers the concept of trigonometric identities, which are fundamental relationships between trigonometric functions.

Exercise 8.2 - This exercise deals with the application of trigonometric ratios to solve real-life problems. Students learn how to use trigonometric ratios to find the height and distance of an object, as well as the angle of elevation and depression. They also learn how to use the Pythagorean theorem to solve problems involving right triangles.

Exercise 8.3 - In this exercise, students learn about complementary angles and how to find the trigonometric ratios of complementary angles. They also learn about the trigonometric ratios of angles greater than 90 degrees and less than 0 degrees, and how to find their values using the reference angle.

Exercise 8.4 - The final exercise covers the concept of trigonometric equations. Students learn how to solve trigonometric equations using the identities and ratios learned in the previous exercises. They also learn how to find the general solution of a trigonometric equation, which involves finding all the possible solutions. Finally, the exercise covers the concept of the period of a trigonometric function and how to find it for different functions.

Access NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry

Exercise 8.1 

1. In $\Delta ABC$ right angled at $B$, $AB=24\text{ cm}$, $BC=7\text{ cm}$. Determine 

(i) $\sin A,\cos A$ 

Ans: Given that in the right angle triangle $\Delta ABC$, $AB=24\text{ cm}$, $BC=7\text{ cm}$.

Let us draw a right triangle $\Delta ABC$, also $AB=24\text{ cm}$, $BC=7\text{ cm}$. We get

(Image will be uploaded soon)

We have to find $\sin A,\cos A$.

We know that for right triangle

$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ and 

$\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$

Now, we need to apply the Pythagoras theorem to find the measure of hypotenuse.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$ 

Here, $AB=24\text{ cm}$, $BC=7\text{ cm}$

We get

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$ 

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( 24 \right)}^{2}}+{{\left( 7 \right)}^{2}}$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}=576+49$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}=625\text{ }c{{m}^{2}}$

$\Rightarrow AC=25\text{ cm}$ 

Now, 

$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ 

$\Rightarrow \sin A=\dfrac{BC}{AC}$ 

$\therefore \sin A=\dfrac{7}{25}$

$\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$

$\Rightarrow \cos A=\dfrac{AB}{AC}$

$\therefore \cos A=\dfrac{24}{25}$

(ii) $\sin C,\cos C$ 

Ans: Given that in the right angle triangle $\Delta ABC$, $AB=24\text{ cm}$, $BC=7\text{ cm}$.

Let us draw a right triangle $\Delta ABC$, also $AB=24\text{ cm}$, $BC=7\text{ cm}$. We get

(Image will be uploaded soon)

We have to find $\sin C,\cos C$.

We know that for right triangle

$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ and 

$\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$

Now, we need to apply the Pythagoras theorem to find the measure of hypotenuse.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$ 

Here, $AB=24\text{ cm}$, $BC=7\text{ cm}$

We get

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$ 

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( 24 \right)}^{2}}+{{\left( 7 \right)}^{2}}$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}=576+49$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}=625\text{ }c{{m}^{2}}$

$\Rightarrow AC=25\text{ cm}$

Now, 

$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ 

$\Rightarrow \sin C=\dfrac{AB}{AC}$ 

$\therefore \sin C=\dfrac{24}{25}$

$\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$

$\Rightarrow \cos C=\dfrac{BC}{AC}$

$\therefore \cos A=\dfrac{7}{25}$

3. In the given figure find $\tan P-\cot R$.

(Image will be uploaded soon)

Ans: Given in the figure, 

$PQ=12\text{ cm}$ 

$PQ=13\text{ cm}$

We know that for right triangle

$\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$ and 

$\cot \theta =\dfrac{\text{adjacent side}}{\text{opposite side}}$

Now, we need to apply the Pythagoras theorem to find the measure of adjacent side/base.

In $\Delta PQR$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$ 

We get

$\Rightarrow {{\left( PR \right)}^{2}}={{\left( PQ \right)}^{2}}+{{\left( QR \right)}^{2}}$ 

$\Rightarrow {{\left( 13 \right)}^{2}}={{\left( 12 \right)}^{2}}+{{\left( QR \right)}^{2}}$

$\Rightarrow 169=144+{{\left( QR \right)}^{2}}$

$\Rightarrow {{\left( QR \right)}^{2}}=169-144$

$\Rightarrow {{\left( QR \right)}^{2}}=25\text{ }c{{m}^{2}}$

$\Rightarrow QR=5\text{ cm}$ 

Now, 

$\tan P=\dfrac{\text{opposite side}}{\text{adjacent side}}$ 

$\Rightarrow \tan P=\dfrac{QR}{PQ}$ 

$\therefore \tan P=\dfrac{5}{12}$

$\cot R=\dfrac{\text{adjacent side}}{\text{opposite side}}$

$\Rightarrow \cot R=\dfrac{QR}{PQ}$

$\therefore \cot R=\dfrac{5}{12}$

$\Rightarrow \tan P-\cot R=\dfrac{5}{12}-\dfrac{5}{12}$

$\therefore \tan P-\cot R=0$ 

3. If $\sin A=\dfrac{3}{4}$, calculate $\cos A$ and $\tan A$.

Ans: Let us consider a right angled triangle $\Delta ABC$. We get

(Image will be uploaded soon)

Given that $\sin A=\dfrac{3}{4}$.

We know that  $\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$.

From the above figure, we get

$\sin A=\dfrac{BC}{AC}$

Therefore, we get

$\Rightarrow BC=3$ and

$\Rightarrow AC=4$ 

Now, we have to find the values of $\cos A$ and $\tan A$.

We know that $\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$ and $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$.

Now, we need to apply the Pythagoras theorem to find the measure of adjacent side/base.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$ 

Here, $AC=4\text{ cm}$, $BC=3\text{ cm}$

We get

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$ 

$\Rightarrow {{4}^{2}}=A{{B}^{2}}+{{3}^{2}}$

$\Rightarrow 16=A{{B}^{2}}+9$

$\Rightarrow A{{B}^{2}}=16-9$

$\Rightarrow A{{B}^{2}}=7$

$\Rightarrow AB=\sqrt{7}\text{ cm}$

Now, we get

$\cos A=\dfrac{AB}{AC}$ 

$\therefore \cos A=\dfrac{\sqrt{7}}{4}$ 

And $\tan A=\dfrac{BC}{AB}$

$\therefore \tan A=\dfrac{3}{\sqrt{7}}$

4. Given $15\cot A=8$. Find $\sin A$ and $\sec A$.

Ans: Let us consider a right angled triangle $\Delta ABC$. We get

(Image will be uploaded soon)

Given that $15\cot A=8$.

We get $\cot A=\dfrac{8}{15}$.

We know that $\cot \theta =\dfrac{\text{adjacent side}}{\text{opposite side}}$.

From the above figure, we get

$\cot A=\dfrac{AB}{BC}$

Therefore, we get

$\Rightarrow BC=15$ and

$\Rightarrow AB=8$ 

Now, we have to find the values of $\sin A$ and $\sec A$.

We know that $\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ and $\sec \theta =\dfrac{\text{hypotenuse}}{\text{adjacent side}}$.

Now, we need to apply the Pythagoras theorem to find the measure of hypotenuse.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$ 

We get

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$ 

$\Rightarrow A{{C}^{2}}={{8}^{2}}+{{15}^{2}}$

$\Rightarrow A{{C}^{2}}=64+225$

$\Rightarrow A{{C}^{2}}=289$

$\Rightarrow AC=17\text{ cm}$

Now, we get

$\sin A=\dfrac{BC}{AC}$ 

$\therefore \sin A=\dfrac{15}{17}$ 

And $\sec A=\dfrac{AC}{AB}$

$\therefore \sec A=\dfrac{17}{8}$

5. Given $\sec \theta =\dfrac{13}{12}$, calculate all other trigonometric ratios.

Ans: Let us consider a right angled triangle $\Delta ABC$. We get

(Image will be uploaded soon)

Given that $\sec \theta =\dfrac{13}{12}$.

We know that $\sec \theta =\dfrac{\text{hypotenuse}}{\text{adjacent side}}$.

From the above figure, we get

$\sec \theta =\dfrac{AC}{AB}$

Therefore, we get

$\Rightarrow AC=13$ and

$\Rightarrow AB=12$ 

Now, we need to apply the Pythagoras theorem to find the measure of the perpendicular/opposite side.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$ 

We get

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$ 

$\Rightarrow {{13}^{2}}={{12}^{2}}+B{{C}^{2}}$

$\Rightarrow 169=144+B{{C}^{2}}$

$\Rightarrow B{{C}^{2}}=25$

$\Rightarrow BC=5\text{ cm}$

Now, we know that

$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$

Here, $\sin \theta =\dfrac{BC}{AC}$ 

$\therefore \sin \theta =\dfrac{5}{13}$ 

We know that $\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$

Here, $\cos \theta =\dfrac{AB}{AC}$ 

$\therefore \cos \theta =\dfrac{12}{13}$

We know that $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$

Here, $\tan \theta =\dfrac{BC}{AB}$ 

$\therefore \tan \theta =\dfrac{5}{12}$

We know that $\operatorname{cosec}\theta =\dfrac{\text{hypotenuse}}{\text{opposite side}}$

Here, $\operatorname{cosec}\theta =\dfrac{AC}{BC}$

$\therefore \operatorname{cosec}\theta =\dfrac{13}{5}$

We know that $\cot \theta =\dfrac{\text{adjacent side}}{\text{opposite side}}$

Here, $\cot \theta =\dfrac{\text{AB}}{BC}$

\[\therefore \cot \theta =\dfrac{12}{5}\] .

6. If $\angle A$ and $\angle B$ are acute angles such that $\cos A=\cos B$, then show that $\angle A=\angle B$.

Ans: Let us consider a right angled triangle $\Delta ABC$. We get

(Image will be uploaded soon)

Given that $\cos A=\cos B$.

In a right triangle $\Delta ABC$, we know that 

$\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$

Here, 

$\cos A=\dfrac{AC}{AB}$ 

And $\cos B=\dfrac{BC}{AB}$ 

As given $\cos A=\cos B$, we get

$\Rightarrow \dfrac{AC}{AB}=\dfrac{BC}{AB}$ 

$\Rightarrow AC=AB$ 

Now, we know that angles opposite to the equal sides are also equal in measure.

Then, we get

$\angle A=\angle B$ 

Hence proved.

7. Evaluate the following if $\cot \theta =\dfrac{7}{8}$

(i) $\dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}$ 

Ans: Let us consider a right angled triangle $\Delta ABC$. We get

(Image will be uploaded soon)

Now, in a right triangle we know that $\cot \theta =\dfrac{\text{adjacent side}}{\text{opposite side}}$.

Here, from the figure $\cot \theta =\dfrac{BC}{AB}$ .

We get

$AB=8$ and 

$BC=7$ 

Now, we need to apply the Pythagoras theorem to find the measure of hypotenuse.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$ 

We get

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$ 

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{8}^{2}}+{{7}^{2}}$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}=64+49$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}=113$

$\Rightarrow AC=\sqrt{113}$ 

Now, we know that 

$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$

Here, we get

$\sin \theta =\dfrac{AB}{AC}=\dfrac{8}{\sqrt{113}}$ and 

$\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$

Here, we get

$\cos \theta =\dfrac{BC}{AC}=\dfrac{7}{\sqrt{113}}$ 

Now, we have to evaluate 

$\dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}$

Applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

\[\Rightarrow \dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}=\dfrac{1-{{\sin }^{2}}\theta }{1-{{\cos }^{2}}\theta }\] 

Substituting the values, we get

\[\Rightarrow \dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}=\dfrac{1-{{\left( \dfrac{8}{\sqrt{113}} \right)}^{2}}}{1-{{\left( \dfrac{7}{\sqrt{113}} \right)}^{2}}}\]

\[\Rightarrow \dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}=\dfrac{1-\dfrac{64}{113}}{1-\dfrac{49}{113}}\]

\[\Rightarrow \dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}=\dfrac{\dfrac{113-64}{113}}{\dfrac{113-49}{113}}\]

\[\Rightarrow \dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}=\dfrac{\dfrac{49}{113}}{\dfrac{64}{113}}\]

\[\therefore \dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}=\dfrac{49}{64}\]

(ii) ${{\cot }^{2}}\theta $ 

Ans: Given that $\cot \theta =\dfrac{7}{8}$.

Now, ${{\cot }^{2}}\theta ={{\left( \dfrac{7}{8} \right)}^{2}}$

$\therefore {{\cot }^{2}}\theta =\dfrac{49}{64}$ 

8. If $3\cot A=4$, check whether $\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A$ or not.

Ans: Let us consider a right angled triangle $\Delta ABC$. We get

(Image will be uploaded soon)

Given that $3\cot A=4$.

We get $\cot A=\dfrac{4}{3}$.

We know that $\cot \theta =\dfrac{\text{adjacent side}}{\text{opposite side}}$.

From the above figure, we get

$\cot A=\dfrac{AB}{BC}$

Therefore, we get

$\Rightarrow BC=3$ and

$\Rightarrow AB=4$ 

Now, we need to apply the Pythagoras theorem to find the measure of hypotenuse.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$ 

We get

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$ 

$\Rightarrow A{{C}^{2}}={{4}^{2}}+{{3}^{2}}$

$\Rightarrow A{{C}^{2}}=16+9$

$\Rightarrow A{{C}^{2}}=25$

$\Rightarrow AC=5$

Now, let us consider LHS of the expression $\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A$, we get

$LHS=\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}$

Now, we know that $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$

Here, we get

$\tan A=\dfrac{BC}{AB}=\dfrac{3}{4}$ 

Substitute the value, we get

$\Rightarrow \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{1-{{\left( \dfrac{3}{4} \right)}^{2}}}{1+{{\left( \dfrac{3}{4} \right)}^{2}}}$

$\Rightarrow \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{1-\dfrac{9}{16}}{1+\dfrac{9}{16}}$

$\Rightarrow \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{\dfrac{16-9}{16}}{\dfrac{16+9}{16}}$

$\Rightarrow \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{\dfrac{7}{16}}{\dfrac{25}{16}}$

$\Rightarrow \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{7}{25}$

Now, let us consider RHS of the expression $\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A$, we get

$RHS={{\cos }^{2}}A-{{\sin }^{2}}A$

We know that $\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ and $\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$.

Here, we get

$\sin A=\dfrac{BC}{AC}=\dfrac{3}{5}$  

And $\cos A=\dfrac{AB}{AC}=\dfrac{4}{5}$

Substitute the values, we get

$\Rightarrow {{\cos }^{2}}A-{{\sin }^{2}}A={{\left( \dfrac{4}{5} \right)}^{2}}-{{\left( \dfrac{3}{5} \right)}^{2}}$

$\Rightarrow {{\cos }^{2}}A-{{\sin }^{2}}A=\dfrac{16}{25}-\dfrac{9}{25}$

$\Rightarrow {{\cos }^{2}}A-{{\sin }^{2}}A=\dfrac{7}{25}$

Hence, we get LHS=RHS

$\therefore \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A$.

9. In $ABC$, right angled at $B$. If $\tan A=\dfrac{1}{\sqrt{3}}$, find the value of 

(i) $\sin A\cos C+\cos A\sin C$ 

Ans: Let us consider a right angled triangle $\Delta ABC$. We get

(Image will be uploaded soon)

Given that $\tan A=\dfrac{1}{\sqrt{3}}$.

In a right triangle, we know that $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$

Here, from the figure we get

$\tan A=\dfrac{BC}{AB}=\dfrac{1}{\sqrt{3}}$

We get $BC=1$ and $AB=\sqrt{3}$ .

Now, we need to apply the Pythagoras theorem to find the measure of hypotenuse.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$ 

We get

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$ 

$\Rightarrow A{{C}^{2}}={{\left( \sqrt{3} \right)}^{2}}+{{1}^{2}}$

$\Rightarrow A{{C}^{2}}=3+1$

$\Rightarrow A{{C}^{2}}=4$

$\Rightarrow AC=2$

We know that $\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ and $\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$.

Here, we get

$\sin A=\dfrac{BC}{AC}=\dfrac{1}{2}$  and $\sin C=\dfrac{AB}{AC}=\dfrac{\sqrt{3}}{2}$

And $\cos A=\dfrac{AB}{AC}=\dfrac{\sqrt{3}}{2}$and $\cos C=\dfrac{BC}{AC}=\dfrac{1}{2}$

Now, we have to find the value of the expression $\sin A\cos C+\cos A\sin C$.

Substituting the values we get

$\Rightarrow \sin A\cos C+\cos A\sin C=\dfrac{1}{2}\times \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}$ 

$\Rightarrow \sin A\cos C+\cos A\sin C=\dfrac{1}{4}+\dfrac{3}{4}$

$\Rightarrow \sin A\cos C+\cos A\sin C=\dfrac{4}{4}$

$\therefore \sin A\cos C+\cos A\sin C=1$

(ii) $\cos A\cos C-\sin A\sin C$

Ans: Let us consider a right angled triangle $\Delta ABC$. We get

(Image will be uploaded soon)

Given that $\tan A=\dfrac{1}{\sqrt{3}}$.

In a right triangle, we know that $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$

Here, from the figure we get

$\tan A=\dfrac{BC}{AB}=\dfrac{1}{\sqrt{3}}$

We get $BC=1$ and $AB=\sqrt{3}$ .

Now, we need to apply the Pythagoras theorem to find the measure of hypotenuse.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$ 

We get

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$ 

$\Rightarrow A{{C}^{2}}={{\left( \sqrt{3} \right)}^{2}}+{{1}^{2}}$

$\Rightarrow A{{C}^{2}}=3+1$

$\Rightarrow A{{C}^{2}}=4$

$\Rightarrow AC=2$

We know that $\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ and $\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$.

Here, we get

$\sin A=\dfrac{BC}{AC}=\dfrac{1}{2}$  and $\sin C=\dfrac{AB}{AC}=\dfrac{\sqrt{3}}{2}$

And $\cos A=\dfrac{AB}{AC}=\dfrac{\sqrt{3}}{2}$and $\cos C=\dfrac{BC}{AC}=\dfrac{1}{2}$

Now, we have to find the value of the expression $\cos A\cos C-\sin A\sin C$.

Substituting the values we get

$\Rightarrow \cos A\cos C-\sin A\sin C=\dfrac{\sqrt{3}}{2}\times \dfrac{1}{2}-\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}$ 

$\Rightarrow \cos A\cos C-\sin A\sin C=\dfrac{\sqrt{3}}{4}-\dfrac{\sqrt{3}}{4}$

$\therefore \Rightarrow \cos A\cos C-\sin A\sin C=0$

10. In $\Delta PQR$, right angled at $Q$, $PR+QR=25\text{ cm}$ and $PQ=5\text{ cm}$. Determine the values of $\sin P,\cos P$ and $\tan P$.

Ans: Let us consider a right angled triangle $\Delta PQR$, we get

(Image will be uploaded soon)

Given that $PR+QR=25\text{ cm}$ and $PQ=5\text{ cm}$.

Let $QR=25-PR$

Now, applying the Pythagoras theorem in $\Delta PQR$, we get

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$ 

We get

$\Rightarrow {{\left( PR \right)}^{2}}={{\left( PQ \right)}^{2}}+{{\left( QR \right)}^{2}}$ 

$\Rightarrow P{{R}^{2}}={{5}^{2}}+{{\left( 25-PR \right)}^{2}}$

$\Rightarrow P{{R}^{2}}=25+{{25}^{2}}+P{{R}^{2}}-50PR$

$\Rightarrow P{{R}^{2}}=P{{R}^{2}}+25+625-50PR$

$\Rightarrow 50PR=650$

$\Rightarrow PR=13\text{ cm}$ 

Therefore, 

$QR=25-13$

$\Rightarrow QR=12\text{ cm}$ 

Now, we know that in right triangle,

$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$, $\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$ and $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$.

Here, we get

$\sin P=\dfrac{QR}{PR}$ 

$\therefore \sin P=\dfrac{12}{13}$ 

$\cos P=\dfrac{PQ}{PR}$ 

$\therefore \cos P=\dfrac{5}{13}$ 

$\tan P=\dfrac{QR}{PQ}$ 

$\therefore \tan P=\dfrac{12}{5}$ 

11. State whether the following are true or false. Justify your answer.

(i) The value of $\tan A$ is always less than $1$.

Ans: The given statement is false. The value of $\tan A$ depends on the length of sides of a right triangle and sides of a triangle may have any measure.

(ii) For some value of angle $A$, $\sec A=\dfrac{12}{5}$.

Ans: We know that in the right triangle $\sec A=\dfrac{\text{hypotenuse}}{\text{adjacent side of }\angle \text{A}}$ .

We know that in the right triangle the hypotenuse is the largest side.

Therefore, the value of $\sec A$ must be greater than $1$.

In the given statement $\sec A=\dfrac{12}{5}$, which is greater than $1$.

Therefore, the given statement is true.

(iii) $\cos A$ is the abbreviation used for the cosecant of angle $A$.

Ans: The given statement is false because $\cos A$ is the abbreviation used for the cosine of angle $A$. Abbreviation used for the cosecant of angle $A$ is $\operatorname{cosec}A$.

(iv) $\cot A$ is the product of $\cot $ and $A$.

Ans: $\cot A$ is the abbreviation used for the cotangent of angle $A$. Hence the given statement is false.

(v) For some angle $\theta $, $\sin \theta =\dfrac{4}{3}$.

Ans: We know that in the right triangle $\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ .

We know that in the right triangle the hypotenuse is the largest side.

Therefore, the value of $\sin \theta $ must be less than $1$.

In the given statement $\sin \theta =\dfrac{4}{3}$, which is greater than $1$.

Therefore, the given statement is false.

Exercise 8.2

1. Evaluate the following:

(i) $\sin 60{}^\circ \cos 30{}^\circ +\sin 30{}^\circ \cos 60{}^\circ $ 

Ans: With the help of trigonometric ratio tables we can find the values of standard trigonometric angles. The trigonometric ratio table is as follows:

We have to evaluate $\sin 60{}^\circ \cos 30{}^\circ +\sin 30{}^\circ \cos 60{}^\circ $.

Substitute the values from the above table, we get

$\Rightarrow \dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\times \dfrac{1}{2}$

$\Rightarrow \dfrac{3}{4}+\dfrac{1}{4}$ 

$\Rightarrow \dfrac{4}{4}$ 

$\therefore \sin 60{}^\circ \cos 30{}^\circ +\sin 30{}^\circ \cos 60{}^\circ =1$.

(ii) $2{{\tan }^{2}}45{}^\circ +{{\cos }^{2}}30{}^\circ -{{\sin }^{2}}60{}^\circ $

Ans: With the help of trigonometric ratio tables we can find the values of standard trigonometric angles. The trigonometric ratio table is as follows:

We have to evaluate $2{{\tan }^{2}}45{}^\circ +{{\cos }^{2}}30{}^\circ -{{\sin }^{2}}60{}^\circ $.

Substitute the values from the above table, we get

$\Rightarrow 2{{\left( 1 \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}$

$\Rightarrow 2+\dfrac{3}{4}-\dfrac{3}{4}$ 

$\Rightarrow 2$ 

$\therefore 2{{\tan }^{2}}45{}^\circ +{{\cos }^{2}}30{}^\circ -{{\sin }^{2}}60{}^\circ =2$.

(iii) $\dfrac{\cos 45{}^\circ }{\sec 30{}^\circ +\operatorname{cosec}30{}^\circ }$

Ans: With the help of trigonometric ratio tables we can find the values of standard trigonometric angles. The trigonometric ratio table is as follows:

We have to evaluate $\dfrac{\cos 45{}^\circ }{\sec 30{}^\circ +\operatorname{cosec}30{}^\circ }$.

Substitute the values from the above table, we get

$\Rightarrow \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2}{\sqrt{3}}+2}$

$\Rightarrow \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2+2\sqrt{3}}{\sqrt{3}}}$ 

\[\Rightarrow \dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2+2\sqrt{3}}\]

Multiplying and dividing by \[\sqrt{3}-1\], we get

\[\Rightarrow \dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2+2\sqrt{3}}\times \dfrac{\sqrt{3}-1}{\sqrt{3}-1}\]

\[\Rightarrow \dfrac{\sqrt{3}\left( \sqrt{3}-1 \right)}{\sqrt{2}\left( 2+2\sqrt{3} \right)\left( \sqrt{3}-1 \right)}\]

\[\Rightarrow \dfrac{\sqrt{3}\left( \sqrt{3}-1 \right)}{2\sqrt{2}\left( \sqrt{3}+1 \right)\left( \sqrt{3}-1 \right)}\]

\[\Rightarrow \dfrac{3-\sqrt{3}}{2\sqrt{2}\left( {{\left( \sqrt{3} \right)}^{2}}-{{1}^{2}} \right)}\]

\[\Rightarrow \dfrac{3-\sqrt{3}}{2\sqrt{2}\left( 3-1 \right)}\]

$\Rightarrow \dfrac{3-\sqrt{3}}{4\sqrt{2}}$ 

$\therefore \dfrac{\cos 45{}^\circ }{\sec 30{}^\circ +\operatorname{cosec}30{}^\circ }=\dfrac{3-\sqrt{3}}{4\sqrt{2}}$

(iv) $\dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\operatorname{cosec}60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ -\cot 45{}^\circ }$

Ans: With the help of trigonometric ratio tables we can find the values of standard trigonometric angles. The trigonometric ratio table is as follows:

We have to evaluate $\dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\operatorname{cosec}60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ -\cot 45{}^\circ }$.

Substitute the values from the above table, we get

$\Rightarrow \dfrac{\dfrac{1}{2}+1-\dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}}+\dfrac{1}{2}+1}$

$\Rightarrow \dfrac{\dfrac{3}{2}-\dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}}+\dfrac{3}{2}}$

$\Rightarrow \dfrac{\dfrac{3\sqrt{3}-4}{2\sqrt{3}}}{\dfrac{3\sqrt{3}+4}{2\sqrt{3}}}$

$\Rightarrow \dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}$

Multiplying and dividing by \[3\sqrt{3}-4\], we get

\[\Rightarrow \dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}\times \dfrac{3\sqrt{3}-4}{3\sqrt{3}-4}\]

Now, applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

\[\Rightarrow \dfrac{{{\left( 3\sqrt{3}-4 \right)}^{2}}}{{{\left( 3\sqrt{3} \right)}^{2}}-{{4}^{2}}}\]

\[\Rightarrow \dfrac{{{\left( 3\sqrt{3}-4 \right)}^{2}}}{{{\left( 3\sqrt{3} \right)}^{2}}-{{4}^{2}}}\]

\[\Rightarrow \dfrac{27+16-24\sqrt{3}}{27-16}\]

$\Rightarrow \dfrac{43-24\sqrt{3}}{11}$ 

$\therefore \dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\operatorname{cosec}60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ -\cot 45{}^\circ }=\dfrac{43-24\sqrt{3}}{11}$

(v) $\dfrac{5{{\cos }^{2}}60{}^\circ +4{{\sec }^{2}}30{}^\circ -{{\tan }^{2}}45{}^\circ }{{{\sec }^{2}}30{}^\circ +{{\cos }^{2}}30{}^\circ }$

Ans: With the help of trigonometric ratio tables we can find the values of standard trigonometric angles. The trigonometric ratio table is as follows:

We have to evaluate $\dfrac{5{{\cos }^{2}}60{}^\circ +4{{\sec }^{2}}30{}^\circ -{{\tan }^{2}}45{}^\circ }{{{\sec }^{2}}30{}^\circ +{{\cos }^{2}}30{}^\circ }$.

Substitute the values from the above table, we get

$\Rightarrow \dfrac{5{{\left( \dfrac{1}{2} \right)}^{2}}+4{{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}}-{{1}^{2}}}{{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}}$

$\Rightarrow \dfrac{5\left( \dfrac{1}{4} \right)+4\left( \dfrac{4}{3} \right)-1}{\left( \dfrac{1}{4} \right)+\left( \dfrac{3}{4} \right)}$

$\Rightarrow \dfrac{\dfrac{15+64-12}{12}}{\dfrac{1+3}{4}}$

$\Rightarrow \dfrac{\dfrac{15+64-12}{12}}{\dfrac{1+3}{4}}$

$\Rightarrow \dfrac{\dfrac{15+64-12}{12}}{\dfrac{4}{4}}$ 

$\Rightarrow \dfrac{\dfrac{67}{12}}{1}$ 

$\therefore \dfrac{5{{\cos }^{2}}60{}^\circ +4{{\sec }^{2}}30{}^\circ -{{\tan }^{2}}45{}^\circ }{{{\sec }^{2}}30{}^\circ +{{\cos }^{2}}30{}^\circ }=\dfrac{67}{12}$.

2. Choose the correct option and justify your choice.

(i) $\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=$ ………

(a) $\sin 60{}^\circ $ 

(b) $\cos 60{}^\circ $ 

(c) $\tan 60{}^\circ $ 

(d) $\sin 30{}^\circ $ 

Ans: The given expression is $\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }$.

We know that from the trigonometric ratio table we have $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$.

Substitute the value in the given expression we get

$\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\dfrac{2\left( \dfrac{1}{\sqrt{3}} \right)}{1+{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\dfrac{\dfrac{2}{\sqrt{3}}}{1+\dfrac{1}{3}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\dfrac{\sqrt{3}}{2}$

From the trigonometric table we know that 

$\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ 

$\cos 60{}^\circ =\dfrac{1}{2}$ 

$\tan 60{}^\circ =\sqrt{3}$ 

$\sin 30{}^\circ =\dfrac{1}{2}$

Hence, $\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\sin 60{}^\circ $.

Therefore, option (A) is the correct answer.

(ii) $\dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=$ ………

(a) $\tan 90{}^\circ $ 

(b) $1$ 

(c) $\sin 45{}^\circ $ 

(d) $0$

Ans: The given expression is $\dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }$.

We know that from the trigonometric ratio table we have $\tan 45{}^\circ =1$.

Substitute the value in the given expression we get

$\dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=\dfrac{1-{{1}^{2}}}{1+{{1}^{2}}}$

$\Rightarrow \dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=\dfrac{1-1}{1+1}$

$\Rightarrow \Rightarrow \dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=\dfrac{0}{2}$

$\therefore \dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=0$

Therefore, option (D) is the correct answer.

(iii) $\sin 2A=2\sin A$ is true when $A=$ ……..

(a) $0{}^\circ $ 

(b) $30{}^\circ $ 

(c) $45{}^\circ $ 

(d) $60{}^\circ $ 

Ans: The given expression is $\sin 2A=2\sin A$.

We know that from the trigonometric ratio table we have 

$\sin 0{}^\circ =0$

$\sin 30{}^\circ =\dfrac{1}{2}$

$\sin 45{}^\circ =\dfrac{1}{\sqrt{2}}$

$\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$  

$\sin 90{}^\circ =1$ 

The given statement is true when $A=0{}^\circ $.

Substitute the value in the given expression we get

$\Rightarrow \sin 2A=2\sin A$

$\Rightarrow \sin 2\times 0{}^\circ =2\sin 0{}^\circ $

$0=0$ 

Therefore, option (A) is the correct answer.

(iv) $\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=$………

(a) $\sin 60{}^\circ $ 

(b) $\cos 60{}^\circ $ 

(c) $\tan 60{}^\circ $ 

(d) $\sin 30{}^\circ $ 

Ans: The given expression is $\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }$.

We know that from the trigonometric ratio table we have $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$.

Substitute the value in the given expression we get

$\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\dfrac{2\left( \dfrac{1}{\sqrt{3}} \right)}{1-{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\dfrac{\dfrac{2}{\sqrt{3}}}{1-\dfrac{1}{3}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\sqrt{3}$

From the trigonometric table we know that 

$\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ 

$\cos 60{}^\circ =\dfrac{1}{2}$ 

$\tan 60{}^\circ =\sqrt{3}$ 

$\sin 30{}^\circ =\dfrac{1}{2}$

Hence, $\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\tan 60{}^\circ $.

Therefore, option (C) is the correct answer.

3. If $\tan \left( A+B \right)=\sqrt{3}$ and $\tan \left( A-B \right)=\dfrac{1}{\sqrt{3}}$, $0{}^\circ <A+B\le 90{}^\circ $. Find $A$ and $B$.

Ans: Given that $\tan \left( A+ dB \right)=\sqrt{3}$ and $\tan \left( A-B \right)=\dfrac{1}{\sqrt{3}}$.

From the trigonometric ratio table we know that $\tan 60{}^\circ =\sqrt{3}$ and $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$.

Then we get

$\tan \left( A+B \right)=\sqrt{3}$

$\Rightarrow \tan \left( A+B \right)=\tan 60{}^\circ $

$\Rightarrow A+B=60{}^\circ $ ……….(1)

Also, $\tan \left( A-B \right)=\dfrac{1}{\sqrt{3}}$

$\Rightarrow \tan \left( A-B \right)=\tan 30{}^\circ $

$\Rightarrow A-B=30{}^\circ $ ……….(2)

Adding eq. (1) and (2), we get

$2A=90{}^\circ $

$\therefore A=45{}^\circ $ 

Substitute the obtained value in eq. (1), we get

$45{}^\circ +B=60{}^\circ $ 

$\Rightarrow B=60{}^\circ -45{}^\circ $ 

$\therefore B=15{}^\circ $ 

Therefore, the values of $A$ and $B$ is $45{}^\circ $ and $15{}^\circ $ respectively.

4. State whether the following are true or false. Justify your answer.

(i) $\sin \left( A+B \right)=\sin A+\sin B$.

Ans: Let us assume $A=30{}^\circ $ and $B=60{}^\circ $.

Now, let us consider LHS of the given expression, we get

$\sin \left( A+B \right)$

Substitute the assumed values in the LHS, we get

$\sin \left( A+B \right)=\sin \left( 30{}^\circ +60{}^\circ  \right)$

$\Rightarrow \sin \left( A+B \right)=\sin \left( 90{}^\circ  \right)$ 

From the trigonometric ratio table we know that $\sin 90{}^\circ =1$, we get

$\Rightarrow \sin \left( A+B \right)=1$

Now, let us consider the RHS of the given expression and substitute the values, we get

$\sin A+\sin B=\sin 30{}^\circ +\sin 60{}^\circ $

From the trigonometric ratio table we know that $\sin 30{}^\circ =\dfrac{1}{2}$ and $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$, we get

$\Rightarrow \sin A+\sin B=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}$

$\Rightarrow \sin A+\sin B=\dfrac{1+\sqrt{3}}{2}$

Thus, $LHS\ne RHS$.

Therefore, the given statement is false.

(ii) The value of $\sin \theta $ increases as $\theta $ increases. 

Ans: The value of sine from the trigonometric ratio table is as follows:

$\sin 0{}^\circ =0$

$\sin 30{}^\circ =\dfrac{1}{2}=0.5$

$\sin 45{}^\circ =\dfrac{1}{\sqrt{2}}=0.707$

$\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}=0.866$  

$\sin 90{}^\circ =1$ 

Therefore, we can conclude that the value of $\sin \theta $ increases as $\theta $ increases. 

Therefore, the given statement is true.

(iii) The value of $\cos \theta $ increases as $\theta $ increases. 

Ans: The value of cosine from the trigonometric ratio table is as follows:

$\cos 0{}^\circ =1$

$\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}=0.866$

$\cos 45{}^\circ =\dfrac{1}{\sqrt{2}}=0.707$

$\cos 60{}^\circ =\dfrac{1}{2}=0.5$  

$\cos 90{}^\circ =0$ 

Therefore, we can conclude that the value of $\cos \theta $ decreases as $\theta $ increases. 

Therefore, the given statement is false.

(iv) \[\sin \theta =\cos \theta \] for all values of \[\theta \].

Ans: The trigonometric ratio table is given as follows:

From the above table we can conclude that \[\sin \theta =\cos \theta \] is true only for $\theta =45{}^\circ $.

\[\sin \theta =\cos \theta \] is not true for all values of $\theta $.

Therefore, the given statement is false.

(v) $\cot A$ is not defined for $A=0{}^\circ $.

Ans: We know that $\cot A=\dfrac{\cos A}{\sin A}$ .

If $A=0{}^\circ $, then $\cot 0{}^\circ =\dfrac{\cos 0{}^\circ }{\sin 0{}^\circ }$

From trigonometric ratio table we get

$\sin 0{}^\circ =0$ and $\cos 0{}^\circ =1$

We get

$\cot 0{}^\circ =\dfrac{1}{0}$, which is undefined.

Therefore, the given statement is true.

Exercise 8.3

1. Evaluate the following:

(i) $\dfrac{\sin 18{}^\circ }{\cos 72{}^\circ }$ 

Ans: The given expression is $\dfrac{\sin 18{}^\circ }{\cos 72{}^\circ }$.

The given expression can be written as \[\dfrac{\sin \left( 90{}^\circ -72{}^\circ  \right)}{\cos 72{}^\circ }\].

Now, we can apply the identity $\sin \left( 90{}^\circ -\theta  \right)=\cos \theta $ , we get

$\dfrac{\sin 18{}^\circ }{\cos 72{}^\circ }=\dfrac{\sin \left( 90{}^\circ -72{}^\circ  \right)}{\cos 72{}^\circ }$

$\Rightarrow \dfrac{\sin 18{}^\circ }{\cos 72{}^\circ }=\dfrac{\cos 72{}^\circ }{\cos 72{}^\circ }$ 

$\therefore \dfrac{\sin 18{}^\circ }{\cos 72{}^\circ }=1$

(ii) $\dfrac{\tan 26{}^\circ }{\cot 64{}^\circ }$ 

Ans: The given expression is $\dfrac{\tan 26{}^\circ }{\cot 64{}^\circ }$.

The given expression can be written as $\dfrac{\tan \left( 90{}^\circ -64{}^\circ  \right)}{\cot 64{}^\circ }$.

Now, we can apply the identity $\tan \left( 90{}^\circ -\theta  \right)=\cot \theta $ , we get

$\dfrac{\tan 26{}^\circ }{\cot 64{}^\circ }=\dfrac{\tan \left( 90{}^\circ -64{}^\circ  \right)}{\cot 64{}^\circ }$

$\Rightarrow \dfrac{\tan 26{}^\circ }{\cot 64{}^\circ }=\dfrac{\cot 64{}^\circ }{\cot 64{}^\circ }$ 

$\therefore \dfrac{\tan 26{}^\circ }{\cot 64{}^\circ }=1$

(iii) $\cos 48{}^\circ -\sin 42{}^\circ $

Ans: The given expression is $\cos 48{}^\circ -\sin 42{}^\circ $.

The given expression can be written as $\cos \left( 90{}^\circ -42{}^\circ  \right)-\sin 42{}^\circ $.

Now, we can apply the identity $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $ , we get

$\cos 48{}^\circ -\sin 42{}^\circ =\cos \left( 90{}^\circ -42{}^\circ  \right)-\sin 42{}^\circ $

$\Rightarrow \cos 48{}^\circ -\sin 42{}^\circ =\sin 42{}^\circ -\sin 42{}^\circ $ 

$\therefore \cos 48{}^\circ -\sin 42{}^\circ =0$

(iv) $cosec 31{}^\circ -\sec 59{}^\circ $

Ans: The given expression is $cosec 31{}^\circ -\sec 59{}^\circ $.

The given expression can be written as $cosec\left( 90{}^\circ -59{}^\circ  \right)-\sec 59{}^\circ $.

Now, we can apply the identity $cosec\left( 90{}^\circ -\theta  \right)=\sec \theta $ , we get

$cosec 31{}^\circ -\sec 59{}^\circ =cosec\left( 90{}^\circ -59{}^\circ  \right)-\sec 59{}^\circ $

$\Rightarrow cosec31{}^\circ -\sec 59{}^\circ =\sec 59{}^\circ -\sec 59{}^\circ $ 

$\therefore cosec 31{}^\circ -\sec 59{}^\circ =0$

2. Show that 

(i) $\tan 48{}^\circ \tan 23{}^\circ \tan 42{}^\circ \tan 67{}^\circ =1$ 

Ans: The given expression is $\tan 48{}^\circ \tan 23{}^\circ \tan 42{}^\circ \tan 67{}^\circ =1$.

Let us consider LHS of the given expression, we get

$\tan 48{}^\circ \tan 23{}^\circ \tan 42{}^\circ \tan 67{}^\circ $

The above expression can be written as 

$\Rightarrow \tan \left( 90{}^\circ -42{}^\circ  \right)\tan \left( 90{}^\circ -67{}^\circ  \right)\tan 42{}^\circ \tan 67{}^\circ $

Now, we can apply the identity $\tan \left( 90{}^\circ -\theta  \right)=\cot \theta $, we get

$\Rightarrow \cot 42{}^\circ \cot 67{}^\circ \tan 42{}^\circ \tan 67{}^\circ $

Now, we know that $\cot A=\dfrac{1}{\tan A}$, we get

$\Rightarrow \dfrac{1}{\tan 42{}^\circ \tan 67{}^\circ }\times \tan 42{}^\circ \tan 67{}^\circ $

$\Rightarrow 1$ 

$\Rightarrow RHS$ 

$\therefore \tan 48{}^\circ \tan 23{}^\circ \tan 42{}^\circ \tan 67{}^\circ =1$ 

(ii) $\cos 38{}^\circ \cos 52{}^\circ -\sin 38{}^\circ \sin 52{}^\circ =0$ 

Ans: The given expression is $\cos 38{}^\circ \cos 52{}^\circ -\sin 38{}^\circ \sin 52{}^\circ =0$.

Let us consider LHS of the given expression, we get

$\cos 38{}^\circ \cos 52{}^\circ -\sin 38{}^\circ \sin 52{}^\circ $

The above expression can be written as 

$\Rightarrow \cos \left( 90{}^\circ -52{}^\circ  \right)\cos \left( 90{}^\circ -38{}^\circ  \right)-\sin 38{}^\circ \sin 52{}^\circ $

Now, we can apply the identity $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $, we get

$\Rightarrow \sin 52{}^\circ \sin 38{}^\circ -\sin 38{}^\circ \sin 52{}^\circ $

$\Rightarrow 0$ 

$\Rightarrow RHS$ 

$\therefore \cos 38{}^\circ \cos 52{}^\circ -\sin 38{}^\circ \sin 52{}^\circ =0$ 

3. Find the value of $A$, if $\tan 2A=\cot \left( A-18{}^\circ  \right)$, where $2A$ is an acute angle.

Ans: Given $\tan 2A=\cot \left( A-18{}^\circ  \right)$……….(1)

Now, we know that $\cot \left( 90{}^\circ -\theta  \right)=\tan \theta $.

Here, we can write $\tan 2A=\cot \left( 90{}^\circ -2A \right)$

Substitute the value in eq. (1), we get

$\Rightarrow \cot \left( 90{}^\circ -2A \right)=\cot \left( A-18{}^\circ  \right)$

Equating both angles, we get

$\Rightarrow \left( 90{}^\circ -2A \right)=\left( A-18{}^\circ  \right)$

$\Rightarrow 90{}^\circ +18{}^\circ =A+2A$ 

$\Rightarrow 108{}^\circ =3A$ 

$\Rightarrow 3A=108{}^\circ $

$\therefore A=36{}^\circ $ 

4. Prove that $A+B=90{}^\circ $, if $\tan A=\cot B$.

Ans: Given that $\tan A=\cot B$.

Now, substitute $\tan A=\cot \left( 90{}^\circ -A \right)$ in the given expression, we get

$\Rightarrow \cot \left( 90{}^\circ -A \right)=\cot B$

Equating both angles, we get

$\Rightarrow \left( 90{}^\circ -A \right)=B$

$\Rightarrow 90{}^\circ =B+A$

$\therefore A+B=90{}^\circ $

Hence proved

5. Find the value of $A$, if $\sec 4A=cosec\left( A-20{}^\circ  \right)$, where $4A$ is an acute angle.

Ans: Given $\sec 4A=cosec\left( A-20{}^\circ  \right)$……….(1)

Now, we know that $cosec\left( 90{}^\circ -\theta  \right)=\sec \theta $.

Here, we can write $\sec 4A=cosec\left( 90{}^\circ -4A \right)$

Substitute the value in eq. (1), we get

$\Rightarrow cosec\left( 90{}^\circ -4A \right)=cosec\left( A-20{}^\circ  \right)$

Equating both angles, we get

$\Rightarrow \left( 90{}^\circ -4A \right)=\left( A-20{}^\circ  \right)$

$\Rightarrow 90{}^\circ +20{}^\circ =A+4A$ 

$\Rightarrow 110{}^\circ =5A$ 

$\Rightarrow 5A=110{}^\circ $

$\therefore A=22{}^\circ $ 

6. If $A,B$ and $C$ are interior angles of a triangle $ABC$, then show that $\sin \left( \dfrac{B+C}{2} \right)=\cos \dfrac{A}{2}$.

Ans: Given that $A,B$ and $C$ are interior angles of a triangle $ABC$.

We know that the sum of interior angles of a triangle is always $180{}^\circ $.

Then, we get

$\Rightarrow \angle A+\angle B+\angle C=180{}^\circ $ 

$\Rightarrow \angle B\angle C=180{}^\circ -\angle A$

Now, divide both sides of the equation by $2$, we get

$\Rightarrow \dfrac{\angle B+\angle C}{2}=\dfrac{180{}^\circ -\angle A}{2}$

$\Rightarrow \dfrac{\angle B+\angle C}{2}=90{}^\circ -\dfrac{\angle A}{2}$

Applying the sine function to the both sides of the equation, we get

$\Rightarrow \sin \left( \dfrac{\angle B+\angle C}{2} \right)=\sin \left( 90{}^\circ -\dfrac{\angle A}{2} \right)$

Now, we know that $\sin \left( 90{}^\circ -\theta  \right)=\cos \theta $.

$\therefore \sin \left( \dfrac{B+C}{2} \right)=\cos \dfrac{A}{2}$.

Hence proved

7. Express $\sin 67{}^\circ +\cos 75{}^\circ $ in terms of trigonometric ratios of angles between $0{}^\circ $ and $45{}^\circ $.

Ans: Given expression $\sin 67{}^\circ +\cos 75{}^\circ $.

Now, we know that $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $.

The given expression can be written as 

$\sin 67{}^\circ +\cos 75{}^\circ =\cos \left( 90{}^\circ -23{}^\circ  \right)+\cos \left( 90{}^\circ -15{}^\circ  \right)$

$\therefore \sin 67{}^\circ +\cos 75{}^\circ =\cos 23{}^\circ +\cos 15{}^\circ $

Therefore, we get the expression in terms of trigonometric ratios of angles between $0{}^\circ $ and $45{}^\circ $.

Exercise 8.4

1. Express the trigonometric ratios $\sin A,\sec A$ and $\tan A$ in terms of $\cot A$.

Ans: For a right triangle we have an identity ${{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A$.

Let us consider the above identity, we get

${{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A$

Now, reciprocating both sides we get

$\Rightarrow \dfrac{1}{{{\operatorname{cosec}}^{2}}A}=\dfrac{1}{1+{{\cot }^{2}}A}$

Now, we know that $\dfrac{1}{{{\operatorname{cosec}}^{2}}A}={{\sin }^{2}}A$, we get

$\Rightarrow {{\sin }^{2}}A=\dfrac{1}{1+{{\cot }^{2}}A}$

$\Rightarrow \sin A=\pm \dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}$

Now, we know that sine value will be negative for angles greater than $180{}^\circ $, for a triangle sine value is always positive with respect to an angle. Then we will consider only positive values.

$\therefore \sin A=\dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}$ 

We know that $\tan A=\dfrac{1}{\cot A}$ 

Also, we will use the identity ${{\sec }^{2}}A=1+{{\tan }^{2}}A$, we get

${{\sec }^{2}}A=1+{{\tan }^{2}}A$

$\Rightarrow {{\sec }^{2}}A=1+\dfrac{1}{{{\cot }^{2}}A}$

$\Rightarrow {{\sec }^{2}}A=\dfrac{{{\cot }^{2}}A+1}{{{\cot }^{2}}A}$

\[\Rightarrow \sec A=\dfrac{\sqrt{{{\cot }^{2}}A+1}}{\sqrt{{{\cot }^{2}}A}}\]

\[\therefore \sec A=\dfrac{\sqrt{{{\cot }^{2}}A+1}}{\cot A}\]

2. Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$.

Ans: 

We know that $\cos A=\dfrac{1}{\sec A}$.

$\therefore \cos A=\dfrac{1}{\sec A}$

For a right triangle we have an identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1$.

Let us consider the above identity, we get

${{\sin }^{2}}A+{{\cos }^{2}}A=1$

Now, we know that $\cos A=\dfrac{1}{\sec A}$, we get

$\Rightarrow {{\sin }^{2}}A=1-{{\cos }^{2}}A$

$\Rightarrow {{\sin }^{2}}A=1-\dfrac{1}{{{\sec }^{2}}A}$

$\Rightarrow \sin A=\sqrt{1-{{\left( \dfrac{1}{\sec A} \right)}^{2}}}$

$\therefore \sin A=\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}$

Also, we will use the identity ${{\sec }^{2}}A=1+{{\tan }^{2}}A$, we get

${{\tan }^{2}}A={{\sec }^{2}}A-1$

$\therefore \tan A=\sqrt{{{\sec }^{2}}A-1}$ 

Now, we know that $\cot A=\dfrac{\cos A}{\sin A}$, we get

$\Rightarrow \cot A=\dfrac{\dfrac{1}{\sec A}}{\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}}$

$\therefore \cot A=\dfrac{1}{\sqrt{{{\sec }^{2}}A-1}}$

We know that $cosecA=\dfrac{1}{\sin A}$, we get

 $\therefore cosecA=\dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}$

3. Evaluate the following:

(i) $\dfrac{{{\sin }^{2}}63{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\cos }^{2}}17{}^\circ +{{\cos }^{2}}73{}^\circ }$ 

Ans: The given expression is $\dfrac{{{\sin }^{2}}63{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\cos }^{2}}17{}^\circ +{{\cos }^{2}}73{}^\circ }$.

The above expression can be written as 

$\dfrac{{{\sin }^{2}}63{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\cos }^{2}}17{}^\circ +{{\cos }^{2}}73{}^\circ }=\dfrac{{{\left[ \sin \left( 90{}^\circ -27{}^\circ  \right) \right]}^{2}}+{{\sin }^{2}}27{}^\circ }{{{\left[ \cos \left( 90{}^\circ -73{}^\circ  \right) \right]}^{2}}+{{\cos }^{2}}73{}^\circ }$

Now, we can apply the identity $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $ and $\sin \left( 90{}^\circ -\theta  \right)=\cos \theta $, we get

$\Rightarrow \dfrac{{{\sin }^{2}}63{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\cos }^{2}}17{}^\circ +{{\cos }^{2}}73{}^\circ }=\dfrac{{{\cos }^{2}}27{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\sin }^{2}}73{}^\circ +{{\cos }^{2}}73{}^\circ }$

Now, by applying the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \dfrac{{{\sin }^{2}}63{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\cos }^{2}}17{}^\circ +{{\cos }^{2}}73{}^\circ }=\dfrac{1}{1}$ 

$\therefore \dfrac{{{\sin }^{2}}63{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\cos }^{2}}17{}^\circ +{{\cos }^{2}}73{}^\circ }=1$  

(ii) $\sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ $ 

Ans: The given expression is $\sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ $ .

The above expression can be written as 

$\sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ =\sin 25{}^\circ \cos \left( 90{}^\circ -25{}^\circ  \right)+\cos 25{}^\circ \sin \left( 90{}^\circ -25{}^\circ  \right)$

Now, we can apply the identity $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $ and $\sin \left( 90{}^\circ -\theta  \right)=\cos \theta $, we get

\[\Rightarrow \sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ =\sin 25{}^\circ \sin 25{}^\circ +\cos 25{}^\circ \cos 25{}^\circ \]

\[\Rightarrow \sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ ={{\sin }^{2}}25{}^\circ +{{\cos }^{2}}25{}^\circ \]

Now, by applying the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

\[\Rightarrow \sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ =1\] 

\[\therefore \sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ =1\]  

4. Choose the correct option and justify your choice:

(i) \[9{{\sec }^{2}}A-9{{\tan }^{2}}A=\] …….

(a) $1$ 

(b) $9$

(c) $8$ 

(d) $0$

Ans: The given expression is $9{{\sec }^{2}}A-9{{\tan }^{2}}A$.

The given expression can be written as 

$\Rightarrow 9{{\sec }^{2}}A-9{{\tan }^{2}}A=9\left( {{\sec }^{2}}A-{{\tan }^{2}}A \right)$

Now, we will use the identity ${{\sec }^{2}}A=1+{{\tan }^{2}}A$, we get

${{\sec }^{2}}A-{{\tan }^{2}}A=1$

$\Rightarrow 9{{\sec }^{2}}A-9{{\tan }^{2}}A=9\left( 1 \right)$

$\therefore 9{{\sec }^{2}}A-9{{\tan }^{2}}A=9$

Therefore, option (B) is the correct answer.

(ii) $\left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)$ 

(a) $0$

(b) $1$

(c) $2$

(d) $-1$

Ans: The given expression is $\left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)$.

We know that the trigonometric functions have values as:

$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$, $cosec\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$

Substituting these values in the given expression, we get

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\left( 1+\dfrac{\sin \theta }{\cos \theta }+\dfrac{1}{\cos \theta } \right)\left( 1+\dfrac{\cos \theta }{\sin \theta }-\dfrac{1}{\sin \theta } \right)$

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\left( \dfrac{\cos \theta +\sin \theta +1}{\cos \theta } \right)\left( \dfrac{\sin \theta +\cos \theta -1}{\sin \theta } \right)$

Now, by applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\dfrac{{{\left( \sin \theta +\cos \theta  \right)}^{2}}-{{1}^{2}}}{\sin \theta \cos \theta }$

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }$

Now, by applying the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\dfrac{1+2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }$

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\dfrac{2\sin \theta \cos \theta }{\sin \theta \cos \theta }$

$\therefore \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=2$

Therefore, option (C) is the correct answer.

(iii) $\left( \sec A+\tan A \right)\left( 1-\sin A \right)=$ ………

(a) $\sec A$ 

(b) $\sin A$ 

(c) $cosecA$ 

(d) $\cos A$ 

Ans: Given expression is $\left( \sec A+\tan A \right)\left( 1-\sin A \right)$.

We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$

Substituting these values in the given expression, we get

$\left( \sec A+\tan A \right)\left( 1-\sin A \right)=\left( \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A} \right)\left( 1-\sin A \right)$

$\Rightarrow \left( \sec A+\tan A \right)\left( 1-\sin A \right)=\left( \dfrac{1+\sin A}{\cos A} \right)\left( 1-\sin A \right)$

$\Rightarrow \left( \sec A+\tan A \right)\left( 1-\sin A \right)=\left( \dfrac{\left( 1+\sin A \right)\left( 1-\sin A \right)}{\cos A} \right)$

Now, by applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

$\Rightarrow \left( \sec A+\tan A \right)\left( 1-\sin A \right)=\left( \dfrac{{{1}^{2}}-{{\sin }^{2}}A}{\cos A} \right)$

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \left( \sec A+\tan A \right)\left( 1-\sin A \right)=\left( \dfrac{{{\cos }^{2}}A}{\cos A} \right)$

$\therefore \left( \sec A+\tan A \right)\left( 1-\sin A \right)=\cos A$

Therefore, option (D) is the correct answer.

(iv) $\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}$ 

(a) ${{\sec }^{2}}A$ 

(b) $-1$ 

(c) ${{\cot }^{2}}A$ 

(d) ${{\tan }^{2}}A$ 

Ans: Given expression is $\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}$.

We know that the trigonometric functions have values as:

$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$.

Substituting these values in the given expression, we get

$\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{1+\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}}{1+\dfrac{{{\cos }^{2}}A}{{{\sin }^{2}}A}}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{\dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A}{{{\cos }^{2}}A}}{\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{{{\sin }^{2}}A}}$

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{\dfrac{1}{{{\cos }^{2}}A}}{\dfrac{1}{{{\sin }^{2}}A}}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}={{\tan }^{2}}A$

Therefore, option (D) is the correct answer.

5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) ${{\left( cosec\theta -cot\theta  \right)}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$ 

Ans: Given expression is ${{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$.

Let us consider the LHS of the given expression, we get

$LHS={{\left( cosec\theta -\cot \theta  \right)}^{2}}$

Now, we know that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$ and $cosec\theta =\dfrac{1}{\sin \theta }$.

By substituting the values, we get

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}={{\left( \dfrac{1}{\sin \theta }-\dfrac{\cos \theta }{\sin \theta } \right)}^{2}}$

 $\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}={{\left( \dfrac{1-\cos \theta }{\sin \theta } \right)}^{2}}$

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{{{\left( 1-\cos \theta  \right)}^{2}}}{{{\sin }^{2}}\theta }$

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{{{\left( 1-\cos \theta  \right)}^{2}}}{1-{{\cos }^{2}}\theta }$

Now, by applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{{{\left( 1-\cos \theta  \right)}^{2}}}{\left( 1-\cos \theta  \right)\left( 1+\cos \theta  \right)}$

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{\left( 1-\cos \theta  \right)}{\left( 1+\cos \theta  \right)}$

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=RHS$

$\therefore {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$

Hence proved

(ii) $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$

Ans: Given expression is $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$.

Let us consider the LHS of the given expression, we get

$LHS=\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}$

Now, taking LCM, we get

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{{{\cos }^{2}}A+\left( 1+\sin A \right)\left( 1+\sin A \right)}{\left( 1+\sin A \right)\cos A}$

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A+2\sin A+1}{\left( 1+\sin A \right)\cos A}$

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{1+2\sin A+1}{\left( 1+\sin A \right)\cos A}$

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2+2\sin A}{\left( 1+\sin A \right)\cos A}$

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2\left( 1+\sin A \right)}{\left( 1+\sin A \right)\cos A}$

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2}{\cos A}$

We know that $sec\theta =\dfrac{1}{\cos \theta }$, we get

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$

\[\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=RHS\]

\[\therefore \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A\]

Hence proved

(iii) $\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta cosec\theta $ 

Ans: Given expression is $\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta $.

Let us consider the LHS of the given expression, we get

$LHS=\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }$

Now, we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$.

By substituting the values, we get

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\left( \dfrac{\dfrac{\sin \theta }{\cos \theta }}{1-\dfrac{\cos \theta }{\sin \theta }}+\dfrac{\dfrac{\cos \theta }{\sin \theta }}{1-\dfrac{\sin \theta }{\cos \theta }} \right)\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\left( \dfrac{\dfrac{\sin \theta }{\cos \theta }}{\dfrac{\sin \theta -\cos \theta }{\sin \theta }}+\dfrac{\dfrac{\cos \theta }{\sin \theta }}{\dfrac{\cos \theta -\sin \theta }{\cos \theta }} \right)\]

 \[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\left( \dfrac{{{\sin }^{2}}\theta }{\cos \theta \left( \sin \theta -\cos \theta  \right)}+\dfrac{{{\cos }^{2}}\theta }{\sin \theta \left( \sin \theta -\cos \theta  \right)} \right)\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\left( \sin \theta -\cos \theta  \right)}\left( \dfrac{{{\sin }^{2}}\theta }{\cos \theta }+\dfrac{{{\cos }^{2}}\theta }{\sin \theta } \right)\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\left( \sin \theta -\cos \theta  \right)}\left( \dfrac{{{\sin }^{3}}\theta -{{\cos }^{3}}\theta }{\sin \theta \cos \theta } \right)\]

Now, by applying the identity \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\], we get

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\left( \sin \theta -\cos \theta  \right)}\left[ \dfrac{\left( \sin \theta -\cos \theta  \right)\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +\sin \theta \cos \theta  \right)}{\sin \theta \cos \theta } \right]\]

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\left( \sin \theta -\cos \theta  \right)}\left[ \dfrac{\left( \sin \theta -\cos \theta  \right)\left( 1+\sin \theta \cos \theta  \right)}{\sin \theta \cos \theta } \right]\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{\left( 1+\sin \theta \cos \theta  \right)}{\sin \theta \cos \theta }\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\sin \theta \cos \theta }+\dfrac{\sin \theta \cos \theta }{\sin \theta \cos \theta }\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\sin \theta \cos \theta }+1\]

We know that $cosec\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$, we get

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\sec \theta cosec\theta +1\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta cosec\theta \]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=RHS\]

\[\therefore \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta cosec\theta \]

Hence proved

(iv) $\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}$ 

Ans: Given expression is $\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}$.

Let us consider the LHS of the given expression, we get

$LHS=\dfrac{1+\sec A}{\sec A}$

Now, we know that $\sec \theta =\dfrac{1}{\cos \theta }$.

By substituting the value, we get

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{1+\dfrac{1}{\cos A}}{\dfrac{1}{\cos A}}\]

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{\dfrac{\cos A+1}{\cos A}}{\dfrac{1}{\cos A}}\]

 \[\Rightarrow \dfrac{1+\sec A}{\sec A}=\cos A+1\]

Multiply and divide by $\left( 1-\cos A \right)$, we get

 \[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{\left( 1+\cos A \right)\left( 1-\cos A \right)}{\left( 1-\cos A \right)}\]

Now, by applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{1-{{\cos }^{2}}A}{\left( 1-\cos A \right)}\]

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{\left( 1-\cos A \right)}\]

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=RHS\]

\[\therefore \dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}\]

Hence proved

(v) $\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=cosecA+\cot A$ 

Ans: Given expression is $\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=cosecA+\cot A$.

Now, let us consider the LHS of the given expression, we get

$LHS=\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}$

Dividing numerator and denominator by $\sin A$, we get

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\dfrac{\cos A}{\sin A}-\dfrac{\sin A}{\sin A}+\dfrac{1}{\sin A}}{\dfrac{\cos A}{\sin A}+\dfrac{\sin A}{\sin A}-\dfrac{1}{\sin A}}$

Now, we know that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$ and $cosec\theta =\dfrac{1}{\sin \theta }$, we get

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\cot A-1+\operatorname{cosec}A}{\cot A+1-\operatorname{cosec}A}$

Now, by applying the identity ${{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A$, substitute $1={{\cot }^{2}}A-{{\operatorname{cosec}}^{2}}A$, we get

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\cot A-\left( {{\cot }^{2}}A-{{\operatorname{cosec}}^{2}}A \right)+\operatorname{cosec}A}{\cot A+{{\cot }^{2}}A-{{\operatorname{cosec}}^{2}}A-\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\cot A-{{\cot }^{2}}A+{{\operatorname{cosec}}^{2}}A+\operatorname{cosec}A}{\cot A+{{\cot }^{2}}A-{{\operatorname{cosec}}^{2}}A-\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{{{\left( \cot A-1+\operatorname{cosec}A \right)}^{2}}}{{{\cot }^{2}}A-1+{{\operatorname{cosec}}^{2}}A+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{{{\left( \cot A-1+\operatorname{cosec}A \right)}^{2}}}{{{\cot }^{2}}A-1+{{\operatorname{cosec}}^{2}}A+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{2{{\operatorname{cosec}}^{2}}A+2\cot A\operatorname{cosec}A-2\cot A-2\operatorname{cosec}A}{{{\cot }^{2}}A-1+{{\operatorname{cosec}}^{2}}A+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{2\operatorname{cosec}A\left( \cot A-\operatorname{cosec}A \right)-2\left( \cot A-\operatorname{cosec}A \right)}{{{\cot }^{2}}A-1+{{\operatorname{cosec}}^{2}}A+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\left( 2\operatorname{cosec}A-2 \right)\left( \cot A-\operatorname{cosec}A \right)}{1-1+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\left( 2\operatorname{cosec}A-2 \right)\left( \cot A-\operatorname{cosec}A \right)}{2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec}A+\cot A$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=RHS$

$\therefore \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec}A+\cot A$ 

Hence proved

(vi) $\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$ 

Ans: Given expression is $\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$.

Let us consider the LHS of the given expression, we get

$LHS=\sqrt{\dfrac{1+\sin A}{1-\sin A}}$

Now, multiply and divide the expression by $\sqrt{1+\sin A}$, we get

$\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sqrt{\dfrac{\left( 1+\sin A \right)\left( 1+\sin A \right)}{\left( 1-\sin A \right)\left( 1+\sin A \right)}}$

Now, by applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sqrt{\dfrac{{{\left( 1+\sin A \right)}^{2}}}{1-{{\sin }^{2}}A}}\]

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\dfrac{1+\sin A}{\sqrt{{{\cos }^{2}}A}}\]

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\dfrac{1+\sin A}{\cos A}\]

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\]

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A\]

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=RHS\]

$\therefore \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$ 

Hence proved

(vii) $\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\tan \theta $ 

Ans: Given expression is $\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\tan \theta $.

Let us consider the LHS of the given expression, we get

$LHS=\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }$

Taking common terms out, we get

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta  \right)}{\cos \theta \left( 2{{\cos }^{2}}\theta -1 \right)}$

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta  \right)}{\cos \theta \left( 2\left( 1-2{{\sin }^{2}}\theta  \right)-1 \right)}$

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta  \right)}{\cos \theta \left( 2-2{{\sin }^{2}}\theta -1 \right)}$

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta  \right)}{\cos \theta \left( 1-2{{\sin }^{2}}\theta  \right)}$

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\dfrac{\sin \theta }{\cos \theta }$

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\tan \theta $

$\Rightarrow \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=RHS$

$\therefore \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\tan \theta $

Hence proved

(viii) ${{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+secA \right)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A$ 

Ans: Given expression is ${{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A$.

Let us consider the LHS of the given expression, we get

$LHS={{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}$

Now, by applying the identity \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\], we get

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}={{\sin }^{2}}A+cosec{{A}^{2}}+2\sin AcosecA+{{\cos }^{2}}A+{{\sec }^{2}}A+2\cos A\sec A\]\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}={{\sin }^{2}}A+{{\cos }^{2}}A+cosec{{A}^{2}}+{{\sec }^{2}}A+2\sin AcosecA+2\cos A\sec A\]We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, $cosec\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$, we get

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=1+cose{{c}^{2}}\theta +{{\sec }^{2}}\theta +2\sin A\dfrac{1}{\sin A}+2\cos A\dfrac{1}{\cos A}\] 

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=1+\left( 1+{{\cot }^{2}}A+1+{{\tan }^{2}}A \right)+2+2\]

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A\]

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=RHS\]

\[\therefore {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A\]

Hence proved

(ix) $\left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}$

Ans: Given expression is $\left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}$.

Let us consider the LHS of the given expression, we get

$LHS=\left( cosecA-\sin A \right)\left( \sec A-\cos A \right)$

We know that $cosec\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$, we get

$\Rightarrow \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\left( \dfrac{1}{\sin A}-\sin A \right)\left( \dfrac{1}{\cos A}-\cos A \right)$

$\Rightarrow \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\left( \dfrac{1-{{\sin }^{2}}A}{\sin A} \right)\left( \dfrac{1-{{\cos }^{2}}A}{\cos A} \right)$

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\left( \dfrac{{{\cos }^{2}}A}{\sin A} \right)\left( \dfrac{{{\sin }^{2}}A}{\cos A} \right)$

$\Rightarrow \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\sin A\cos A$

Now, consider the RHS of the given expression, we get

$RHS=\dfrac{1}{\tan A+\cot A}$

Now, we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$.

$\Rightarrow \dfrac{1}{\tan A+\cot A}=\dfrac{1}{\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}}$

$\Rightarrow \dfrac{1}{\tan A+\cot A}=\dfrac{1}{\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\sin A\cos A}}$

$\Rightarrow \dfrac{1}{\tan A+\cot A}=\dfrac{\sin A\cos A}{{{\sin }^{2}}A+{{\cos }^{2}}A}$

Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \dfrac{1}{\tan A+\cot A}=\sin A\cos A$

Here, we get LHS=RHS

$\therefore \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}$ 

Hence proved

(x) $\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$

Ans: Given expression is $\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$.

Let us consider the LHS of the given expression, we get

$LHS=\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}$

By applying the identities ${{\sec }^{2}}A=1+{{\tan }^{2}}A$ and ${{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A$, we get

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{se{{c}^{2}}A}{{{\operatorname{cosec}}^{2}}A}$

We know that $cosec\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$, we get

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{\dfrac{1}{{{\cos }^{2}}A}}{\dfrac{1}{{{\sin }^{2}}A}}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}={{\tan }^{2}}A$

Now, consider the RHS of the given expression, we get

$RHS={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$

Now, we know that $\cot \theta =\dfrac{1}{\tan \theta }$, we get

$\Rightarrow {{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\left( \dfrac{1-\tan A}{1-\dfrac{1}{\tan A}} \right)}^{2}}$

$\Rightarrow {{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\left( \dfrac{1-\tan A}{\dfrac{\tan A-1}{\tan A}} \right)}^{2}}$

$\Rightarrow {{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\left( -\tan A \right)}^{2}}$

$\Rightarrow {{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A$

Here, we get LHS=RHS

$\therefore \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$ 

Hence proved

NCERT Solutions for Class 10 Maths Chapter 8 - Free PDF Download

We at Vedantu will help students overcome all difficulties and get the best possible marks in their examinations. We provide students with a PDF of NCERT questions that are solved. All these exercise questions will allow them to understand the kinds of sums frequently asked in the examination.

You can find the Solutions of All the Class 10 Maths Chapters below.

NCERT Solutions for Class 10 Maths

• Chapter 1 - Real Numbers

• Chapter 2 - Polynomials

• Chapter 3 - Pair of Linear Equations in Two Variables

• Chapter 4 - Quadratic Equations

• Chapter 5 - Arithmetic Progressions

• Chapter 6 - Triangles

• Chapter 7 - Coordinate Geometry

• Chapter 8 - Introduction to Trigonometry

• Chapter 9 - Some Applications of Trigonometry

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Areas Related to Circles

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

About the Chapter - Introduction to Trigonometry

Trigonometry is all about triangles. It is all about right-angled triangles, triangles with one angle equal to 90 degrees, to be more precise. It's a method that helps us find a triangle's missing angles and missing sides. The ‘trigono’ word means triangle and the ‘metry’ word means to measure. 

Chapter 8 Introduction to Class 10 Trigonometry NCERT Syllabus is divided into five parts and four exercises. The last part of the exercise consists of problems that can be pictured using the right angle triangle. The second section consists of an introduction to trigonometric ratios with examples. It is accompanied by an exercise at the end of the section and the derivation of sine, cosine, and other trigonometric functions.  

The third part consists of trigonometric ratios about measurement. The fourth section of Chapter 8 Class 10 Maths consists of a few solved examples, trigonometric ratio criteria for complementary angles, and an exercise. In the introduction to Trigonometry Class 10, the fifth section consists of the subject relating to trigonometric identities, with a few examples, and ends with an exercise.

Key Concepts at a Glance

The following are the key concepts that are covered in the chapter.

Opposite & Adjacent Sides in a Right Angled Triangle

• The sides of a right-angle triangle are referred to as opposite, adjacent, and hypotenuse.

• The hypotenuse is the side that is opposite to the right angle. It's also the triangle's longest side.

• Adjacent and opposite sides are in respect to the angle specified and might vary accordingly. The opposite side is the side that is opposite the specified angle, while the adjacent side is the side that is right by the specified angle.

Basic Trigonometric Ratios

In ΔABC, right-angled at ∠B, the trigonometric ratios of the ∠A are as follows:

• sin A=opposite side/hypotenuse=BC/AC

• cos A=adjacent side/hypotenuse=AB/AC

• tan A=opposite side/adjacent side=BC/AB

• cosec A=hypotenuse/opposite side=AC/BC

• sec A=hypotenuse/adjacent side=AC/AB

• cot A=adjacent side/opposite side=AB/BC

Standard Values of Trigonometric Ratios

Complementary Trigonometric Ratios:

• sin (90o− θ) = cos θ

• cos (90o− θ) = sin θ

• tan (90o− θ) = cot θ

• cot (90o− θ) = tan θ

• cosec (90o− θ) = sec θ

• sec (90o− θ) = cosec θ

Benefits of NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 10 Maths provides many benefits. Some of the benefits and features of these NCERT Strategies are:

• To save students from the needless burden on their minds, complicated solutions are broken down into simple sections.

• Solutions are written in a lucid language to make it easier for students to understand. 

• The solutions are kept concise, short, and self-explanatory. 

• To facilitate the understanding of the concept, topics and answers are incorporated with relevant illustrations. 

• All the solutions are provided in adherence to the guidelines set by CBSE. 

• In the chapter Introduction to Trigonometry, these NCERT Solutions are precisely written so that students do not face any difficulties while solving any questions. 

• Our experts have made sure that it is straightforward to understand these NCERT solutions. 

• In the form of solutions, the PDF encompasses the entire syllabus and theory.

• The solutions are provided with the new syllabus and exam requirements in mind.

All study materials, and NCERT solutions are available on the website and app of Vedantu. All the NCERT solutions are available for free download on Vedantu in PDF format. Along with this, students can also view additional study materials provided by Vedantu, for Class 10 Maths 

• CBSE Class 10 Maths Chapter Wise Formulas

• CBSE Class 10 Maths Revision Notes

• CBSE Class 10 Maths Syllabus

This was the complete discussion on Class 10 Maths Chapter 8 Introduction to Trigonometry. We hope to have helped you solve your queries. Download the NCERT solutions now, to ace your Class 10 examination