NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry (Ex 8.2) Exercise 8.2

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry (Ex 8.2) Exercise 8.2

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Access NCERT Solutions for Class -10 Maths  Chapter 8 – Introduction to Trigonometry part-1

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

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:

Exact Values of Trigonometric Functions

Angle $\theta $ 

$\sin \theta $ 

$\cos \theta $ 

$\tan \theta $ 

Degrees

Radians

$0{}^\circ $ 

$0$ 

$0$

$1$ 

$0$

$30{}^\circ $ 

$\dfrac{\pi }{6}$ 

$\dfrac{1}{2}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{\sqrt{3}}$ 

$45{}^\circ $ 

$\dfrac{\pi }{4}$ 

$\dfrac{1}{\sqrt{2}}$

$\dfrac{1}{\sqrt{2}}$ 

$1$ 

$60{}^\circ $ 

$\dfrac{\pi }{3}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{2}$

$\sqrt{3}$

$90{}^\circ $ 

$\dfrac{\pi }{2}$ 

$1$ 

$0$

Not defined 


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:

Exact Values of Trigonometric Functions

Angle $\theta $ 

$\sin \theta $ 

$\cos \theta $ 

$\tan \theta $ 

Degrees

Radians

$0{}^\circ $ 

$0$ 

$0$

$1$ 

$0$

$30{}^\circ $ 

$\dfrac{\pi }{6}$ 

$\dfrac{1}{2}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{\sqrt{3}}$ 

$45{}^\circ $ 

$\dfrac{\pi }{4}$ 

$\dfrac{1}{\sqrt{2}}$

$\dfrac{1}{\sqrt{2}}$ 

$1$ 

$60{}^\circ $ 

$\dfrac{\pi }{3}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{2}$

$\sqrt{3}$

$90{}^\circ $ 

$\dfrac{\pi }{2}$ 

$1$ 

$0$

Not defined 


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:

Exact Values of Trigonometric Functions

Angle $\theta $ 

$\sin \theta $ 

$\cos \theta $ 

$\tan \theta $ 

Degrees

Radians

$0{}^\circ $ 

$0$ 

$0$

$1$ 

$0$

$30{}^\circ $ 

$\dfrac{\pi }{6}$ 

$\dfrac{1}{2}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{\sqrt{3}}$ 

$45{}^\circ $ 

$\dfrac{\pi }{4}$ 

$\dfrac{1}{\sqrt{2}}$

$\dfrac{1}{\sqrt{2}}$ 

$1$ 

$60{}^\circ $ 

$\dfrac{\pi }{3}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{2}$

$\sqrt{3}$

$90{}^\circ $ 

$\dfrac{\pi }{2}$ 

$1$ 

$0$

Not defined 


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:

Exact Values of Trigonometric Functions

Angle $\theta $ 

$\sin \theta $ 

$\cos \theta $ 

$\tan \theta $ 

Degrees

Radians

$0{}^\circ $ 

$0$ 

$0$

$1$ 

$0$

$30{}^\circ $ 

$\dfrac{\pi }{6}$ 

$\dfrac{1}{2}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{\sqrt{3}}$ 

$45{}^\circ $ 

$\dfrac{\pi }{4}$ 

$\dfrac{1}{\sqrt{2}}$

$\dfrac{1}{\sqrt{2}}$ 

$1$ 

$60{}^\circ $ 

$\dfrac{\pi }{3}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{2}$

$\sqrt{3}$

$90{}^\circ $ 

$\dfrac{\pi }{2}$ 

$1$ 

$0$

Not defined 


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:

Exact Values of Trigonometric Functions

Angle $\theta $ 

$\sin \theta $ 

$\cos \theta $ 

$\tan \theta $ 

Degrees

Radians

$0{}^\circ $ 

$0$ 

$0$

$1$ 

$0$

$30{}^\circ $ 

$\dfrac{\pi }{6}$ 

$\dfrac{1}{2}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{\sqrt{3}}$ 

$45{}^\circ $ 

$\dfrac{\pi }{4}$ 

$\dfrac{1}{\sqrt{2}}$

$\dfrac{1}{\sqrt{2}}$ 

$1$ 

$60{}^\circ $ 

$\dfrac{\pi }{3}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{2}$

$\sqrt{3}$

$90{}^\circ $ 

$\dfrac{\pi }{2}$ 

$1$ 

$0$

Not defined 


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

  1. $\sin 60{}^\circ $ 

  2. $\cos 60{}^\circ $ 

  3. $\tan 60{}^\circ $ 

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

  1. $\tan 90{}^\circ $ 

  2. $1$ 

  3. $\sin 45{}^\circ $ 

  4. $0$

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

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

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

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

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

Therefore, option (D) is the correct answer.


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

  1. $0{}^\circ $ 

  2. $30{}^\circ $ 

  3. $45{}^\circ $ 

  4. $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)  $\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=$………

  1. $\sin 60{}^\circ $ 

  2. $\cos 60{}^\circ $ 

  3. $\tan 60{}^\circ $ 

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

Exact Values of Trigonometric Functions

Angle $\theta $ 

$\sin \theta $ 

$\cos \theta $ 

$\tan \theta $ 

Degrees

Radians

$0{}^\circ $ 

$0$ 

$0$

$1$ 

$0$

$30{}^\circ $ 

$\dfrac{\pi }{6}$ 

$\dfrac{1}{2}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{\sqrt{3}}$ 

$45{}^\circ $ 

$\dfrac{\pi }{4}$ 

$\dfrac{1}{\sqrt{2}}$

$\dfrac{1}{\sqrt{2}}$ 

$1$ 

$60{}^\circ $ 

$\dfrac{\pi }{3}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{2}$

$\sqrt{3}$

$90{}^\circ $ 

$\dfrac{\pi }{2}$ 

$1$ 

$0$

Not defined 

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.


NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.2

Opting for the NCERT solutions for Ex 8.2 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 8.2 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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FAQs (Frequently Asked Questions)

Q1. What are the topics/ sub-topics covered in this Class 10 Maths Chapter 8 Exercise 8.2?

Ans. NCERT Solutions for Class 10 Maths Chapter 8 deals with Introduction to Trigonometry. The topics/ sub-topics covered in this Class 10 Maths Chapter 8 Exercise 8 are: 

  • Introduction to Trigonometry

  • Introduction

  • Trigonometric Ratios

  • Trigonometric Ratios of Some Specific Angles

  • Trigonometric Ratios of Complementary Angles

  • Trigonometric Identities

  • Summary

Q2. How many questions are there Class 10 Maths Chapter 8 Exercise 8.2 of NCERT textbook?

Ans. Class 10 Maths Chapter 8 Exercise 8.2 of NCERT textbook contains four questions in total. Out of which, one was short answer type, two long answer type, and one multiple choice question. Answer to these questions have also been provided by Vedantu.

Q3. What are the benefits of using NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.2?

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Q4. Can we download NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.2 for free?

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Q6. How much time is needed to complete Exercise 8.2 of Class 10?

Ans: Well, it does not take much time to complete Exercise 8.2 of Class 10. What is needed the most is ‘practice’. If you have perfect practice and speed, you can easily solve it in 1-2 days. Again, it wholly depends on the student’s capability. If you devote at least 2 hours per day, you can solve it faster. A good solving technique is equally important for the completion of the exercise.

Q7. Where can I get solutions for Chapter Trigonometry?

Ans: The solutions for Chapter Trignometry are available on Vedantu. This platform has been helping students for a long time. It is one of the trusted study guides. The questions and answers are prepared by the experts keeping the CBSE guidelines in mind. Also, the solutions are available free of cost. Trigonometry is one of the important chapters of Class 8. Therefore, proper efforts are vital. 

Q8. How to score maximum in Exercise 8.2 of Class 10 Chapter 8 Maths?

Ans: To get maximum marks in Exercise 8.2 of Class 10 Chapter 8 Maths, revise at regular intervals. It is a time-consuming exercise, so give enough hours for understanding. Don’t just do it in a hurry, as there are chances that you might miss the important steps. The key to scoring good marks is practising, and if this is proper then no one can stop you from getting maximum marks.

Q9. Do examples of Exercise 8.2 of Class 10 Chapter 8 Maths also come in board exams?

Ans: Yes. Examples of Exercise 8.2 of Class 10 Chapter 8 Maths can also come in board exams. It has happened many times that the question comes from the examples, and only the numbers are replaced. Apart from the digits, the entire question is the same. So, you should be familiar with the examples too. Examples do help in solving the exercise.

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