Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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

Last updated date: 11th Sep 2024
Total views: 616.8k
Views today: 18.16k

## Exercise 8.2 Class 10 Maths NCERT Solutions for Chapter 8 - Free PDF Download

NCERT Solutions for Exercise 8.2 Class 10 Maths, Chapter 8, Introduction to Trigonometry, are available here with all the solved problems. It is available in free PDF format, which students can download and study with ease. The NCERT solutions are designed and reviewed by our subject matter experts with suitable diagrams, and the solutions are solved in a step-by-step manner. The solutions were created in accordance with the latest NCERT syllabus and guidelines of the CBSE board. All the solutions are given chapter-wise for easy access, helping students to solve the problems easily. The NCERT Class 10 Exercise 8.2 Maths Solutions for all the chapters can be downloaded at Vedantu for free.

Table of Content
1. Exercise 8.2 Class 10 Maths NCERT Solutions for Chapter 8 - Free PDF Download
2. Glance on NCERT Solutions Maths Chapter 8 Ex 8.2 Class 10 | Vedantu
3. Access NCERT Solutions for Class -10 Maths Chapter 8 – Introduction to Trigonometry
4. NCERT Solutions for Class 10 Maths Chapter 8 Exercises
5. CBSE Class 10 Maths Chapter 8 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs

## Glance on NCERT Solutions Maths Chapter 8 Ex 8.2 Class 10 | Vedantu

• This chapter focuses on the formula of foundational trigonometric identities.

• This chapter includes primarily theoretical questions that involve proving various identities and demonstrating the interrelationships between different trigonometric functions.

• This chapter of Chapter 8 Maths Exercise 8.2 Class 10 helps students understand  Introduction to Trigonometry.

• There are links to video tutorials explaining Chapter 8 Ex 8.2 Class 10 Introduction to Trigonometry for better understanding.

• There are three exercises (19 fully solved questions) in Class 10 Maths, Chapter 8, Exercise 8.2  Introduction to Trigonometry.

Competitive Exams after 12th Science
Watch videos on
NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry - Exercise 8.2
TRIGONOMETRY Class 10 in One Shot (Complete Chapter) | CBSE 10 Math Chapter 8 [Term 1 Exam] Vedantu
Vedantu 9&10
Subscribe
Share
7.2K likes
155.2K Views
3 years ago
TRIGONOMETRY L-2 (Trigonometric Identities) CBSE Class 10 Maths Chapter 8 | Term 1 Exam | Vedantu
Vedantu 9&10
8K likes
141.9K Views
3 years ago
Play Quiz
TRIGONOMETRY L-1 [Introduction and Trigonometric Ratios of Some Specific Angles] CBSE 10 Math Chap 8
Vedantu 9&10
13.5K likes
269.6K Views
3 years ago
Play Quiz
Trigonometry in One Shot | CBSE Class 10 Maths Chapter 8 | NCERT Solutions | Vedantu Class 9 and 10
Vedantu 9&10
7.7K likes
183.4K Views
3 years ago
See More

## 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.

## Conclusion

For a thorough understanding of trigonometry fundamentals, NCERT Solutions for Ex 8.2 Class 10 Maths Chapter 8, "Introduction to Trigonometry," is an excellent resource. This chapter is essential because it establishes a number of ideas that are fundamental to further research in engineering, physics, and mathematics. In particular, students should concentrate on becoming skilled in trigonometric identities and ratios, as well as how to use them to solve problems that include heights and distances. It is necessary to fully understand these ideas in order to answer a variety of exam questions.

## NCERT Solutions for Class 10 Maths Chapter 8 Exercises

 Exercise Number of Questions Exercise 8.1 11 Questions and Solutions Exercise 8.3 4 Questions and Solutions

## Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry - Exercise 8.2

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

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

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

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.

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

Vedantu NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.2 are extremely helpful for all the students as these are designed by the Maths experts that cover all the questions from the textbook. These NCERT Solutions are created as per the latest CBSE syllabus and guidelines, in accordance with the exam pattern prescribed by the CBSE.

These NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.2 provide a strong foundation for every concept. Students can clarify their doubts and understand the fundamentals of the chapter through these solutions. With the help of NCERT Solutions for Class 10, it becomes easier to solve the difficult problems in each exercise.

Yes, you can download NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.2 for free of cost from Vedantu official website i.e. vedantu.com. You can download these solutions from Vedantu mobile app as well. These solutions are developed by Vedantu’s subject matter experts who hold years of experience in the respective fields. Hence these solutions are accurate and can be downloaded without any hassle.

5. What should I focus on in exercise 8.2 class 10 for my exams?

In class 10 maths ch 8 ex 8.2 focus on mastering the derivation and application of trigonometric ratios for standard angles. Ensure you understand how these ratios are calculated and can apply them confidently to solve problems. This foundational knowledge is crucial for tackling more complex trigonometry questions and applications you will encounter in higher studies.

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

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.

7. Where can I get solutions for Chapter Trigonometry?

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.

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

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.

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

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.

10. Are there any tricky problems in Class 10 Ex 8.2 I should be aware of?

In class 10 maths ex 8.2 , watch out for problems that involve angles beyond the standard 0°, 30°, 45°, 60°, and 90°. These might require you to use trigonometric identities or complementary angle relationships to find solutions. Understanding how to handle these non-standard angles is crucial for solving such problems accurately and efficiently in your exams.