RD Sharma Class 12 Exercise 3.10 Inverse Trigonometric Functions Solutions Maths

Q: Why should I refer to the RD Sharma Class 12 Maths Book?

The RD Sharma Class 12 Maths Book is the best book for exam preparation. If you want to take competitive exams like IIT JEE, this book will help you get there. RD Sharma Class 12 Maths Solutions Chapter 3 Exercise 3.5 contains everything you need to know about the subject, including examples, tips, tricks, and practice questions.

Q: Are these solutions free and up to date?

Yes, Career360's solutions are free and up to date with the most recent book questions. You can take advantage of the benefits by visiting the website. The high quality of these solutions will make you feel like a knowledge elite.

Q: What are the topics covered in Inverse Trigonometry?

The chapter 'Inverse Trigonometry' involves inverses of cosine, sine, tangent, cotangent, secant, and cosecant functions.

Q: Is the chapter Inverse Trigonometry hard to learn?

The students are capable of learning any topics with a good amount of practice. Even though Inverse trigonometry is complex, the RD Sharma Class 12 Chapter 3 Exercise 3.10 will lend a helping hand to the students.

Q: What makes the RD Sharma solution books stand out from the other books available in the market?

The RD Sharma books have lots of practice problems that gives the students a strong foundation regarding the concepts. Therefore, many people use it making it one of the most trusted books of the class 12 students.

RD Sharma Class 12 Exercise 3.10 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 11:32 AM IST

RD Sharma books have been popular for a very long time due to the quality and coverage of every topic in the best way. Mathematics is tough for many students, but with RD Sharma Class 12th Exercise 3.10 Solution, any student can learn the concepts easily.

RD Sharma Class 12th Exercise 3.10, Inverse Trigonometry involves inverses of cosine, sine, tangent, cotangent, secant, and cosecant functions. These topics are questioned individually to clear the concept and solve complicated problems without any difficulty. Not every child would feel these topics easy to workout. The way all the sums are explained in this book are effective for the students and also simple for the teacher to teach the concepts. RD Sharma Class 12th Exercise 3.10 involves 14 questions that are formulated considering the latest trends of the competition and board exams.

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RD Sharma Class 12 Solutions Chapter 3 Inverse Trigonometric Functions - Other Exercise

Inverse Trigonometric Functions Excercise: 3.10

Inverse Trigonometric Function Exercise 3.10 Question 1 (i)

Answer: $0$
Hint: Check if there is any relation between $\sec ^{-1}x$ and $\cos^{-1}x$ . Use inverse trigonometric functions properties to solve. $( \therefore \sec^{-1}x = \cos^{-1}\frac{1}{x} )$
Given: $\cot\left (\sin^{-1}\frac{3}{4}+ \sec^{-1}\frac{4}{3} \right )$
Solution: Let replace $\sec^{-1}\frac{4}{3}$ by $\cos^{-1}\frac{3}{4}$ because
$\therefore \sec^{-1}x = \cos^{-1}\frac{1}{x}$
$= \cos^{-1}\frac{3}{4}$
Now, $\Rightarrow \cot\left (\sin^{-1}\frac{3}{4}+ \sec^{-1}\frac{4}{3} \right )$
Again by property, $\sin^{-1}x+ \cos^{-1}x= \frac{\pi }{2}$
Now $\Rightarrow \cot\left (\sin^{-1}\frac{3}{4}+ \cos^{-1}\frac{3}{4} \right )$
$\! \! \! \! \! \! \! \! \! \Rightarrow \cot\frac{\pi }{2}\\ = 0 \; \; \; \; \; \; [\because \cot\frac{\pi }{2}=0]$
Hence, $\cot\left (\sin^{-1}\frac{3}{4}+ \sec^{-1}\frac{4}{3} \right )= 0$
Concept: Properties and relations between inverse trigonometric functions.
Note: Inverse trigonometric functions remember relation between all trigonometric functions. Also, try to remember value of trigonometric functions.

Inverse Trigonometric Function Exercise 3.10 Question 1 (ii)

Answer:

Answer: $-1$
Hint: Check if there is any relation between $\sec ^{-1}x$ and $\cos^{-1}x$ . Use inverse trigonometric functions properties to solve. $( \therefore \tan^{-1}\frac{1}{x} = - \pi + \cot^{-1}x )$
Given: $\sin\left (\tan^{-1}x+ \tan^{-1}\frac{1}{x} \right ) f\! or\: x < 0$
Solution: Let replace $\tan^{-1}\frac{1}{x}$ by $(- \pi + \cot^{-1}x) , x < 0$
Now, $\sin\left (\tan^{-1}x+ \tan^{-1}\frac{1}{x} \right )$
$\Rightarrow \sin\left (\tan^{-1}x- \pi + \cot^{-1}x \right )$
Again by property, $\tan^{-1}x+ \cot^{-1}x = \frac{\pi }{2}$
$\! \! \! \! \! \! \! \! \! \! = \sin\frac{\pi }{2}- \pi \\ = \sin-\frac{\pi }{2}\\ = - 1$
Concept: Properties and relations between inverse trigonometric functions.
Note: Inverse trigonometric functions remember relation between all trigonometric functions. Also, try to remember value of trigonometric functions.

Inverse Trigonometric Function Exercise 3.10 Question 1 (iii)

Answer:

Answer: 1
Hint: Check if there is any relation between $\sec ^{-1}x$ and $\cos^{-1}x$ . Use inverse trigonometric functions properties to solve. $( \therefore \tan^{-1}\frac{1}{x} = \cot^{-1}x )$
Given: $\sin\left (\tan^{-1}x+ \tan^{-1}\frac{1}{x} \right )\: f\! or\: x> 0$
Solution: Let replace $\tan^{-1}\frac{1}{x}$ by $( \cot^{-1}x) , x < 0$
Now, $\Rightarrow \sin\left (\tan^{-1}x+ \tan^{-1}\frac{1}{x} \right )$
$\Rightarrow \sin\left (\tan^{-1}x+ \cot^{-1}x \right )$
Again by property, $\tan^{-1}x + \cot^{-1}x= \frac{\pi }{2}$
$= \sin\left ( \frac{\pi }{2} \right )$
$= \sin\frac{\pi }{2}$
$= 1$
Concept: Properties and relations between inverse trigonometric functions.
Note: Inverse trigonometric functions remember relation between all trigonometric functions. Also, try to remember value of trigonometric functions.

Inverse Trigonometric Function Exercise 3.10 Question 1 (iv)

Answer: $0$
Hint: Check properties of trigonometric functions
Given: $\cot\left ( \tan^{-1}a+\cot^{-1}a \right ) f\! or\: x < 0$
Solution:
We know that, $\tan^{-1}x +\cot^{-1} x= \frac{\pi }{2}$
Therefore $\Rightarrow \cot\left ( \tan^{-1}a+\cot^{-1}a \right )$
$\Rightarrow \cot = \frac{\pi }{2}$
$= 0\! \: \: \: \: \: \: \left [ \frac{\pi }{2}\: \epsilon \: \left [ 0,\pi \right ] \right ]$
Concept: Properties and relations between inverse trigonometric functions.
Note: Deriving value of other trigonometric functions from basic sine and cosine functions.

Inverse Trigonometric Function Exercise 3.10 Question 1 (v)

Answer:

Answer: $0$
Hint: Check properties of trigonometric functions
Given: $\cos\left ( \sec ^{-1} x+ cosec^{-1}x\right ) , |x| \geq 1$
Solution:
We know that, $\sec^{-1}x+ cosec^{-1}x=\frac{\pi }{2}$
Therefore $\Rightarrow \cos\left ( \sec ^{-1} x+ cosec^{-1}x\right )$
$\Rightarrow \cos \frac{\pi }{2}= 0$
$= 0$
Concept: Properties and relations between inverse trigonometric functions.
Note: Deriving value of other trigonometric functions from basic sine and cosine functions.

Inverse Trigonometric Function Exercise 3.10 Question 2

Answer:

Answer: $\frac{3\pi }{4}$
Hint: Check how $\sin^{-1}x$ can be expressed in terms of $\cos^{-1}x$
Given: $\cos^{-1}x+ \cos^{-1}y =\frac{\pi }{4}$
Here, we have to compute
$\sin^{-1}x+ \sin^{-1}y$
Solution:
Let’s replace $\sin^{-1}x$ by $\frac{\pi }{2} - \cos^{-1}x$
Now, $\sin^{-1}x+ \sin^{-1}y$
$\! \! \! \! \! \! \! \! \! \! \Rightarrow \frac{\pi }{2} - \cos^{-1}x + \frac{\pi }{2} - \cos^{-1}y\\ = \frac{\pi }{2} + \frac{\pi }{2} - \cos^{-1}x+ \cos^{-1}y\\ = \pi - \frac{\pi }{4}\\ = \frac{3\pi }{4}$
Concept: Properties and relations between inverse trigonometric functions.
Note: Remember properties of Trigonometric functions.

Inverse Trigonometric Function Exercise 3.10 Question 3

Answer: $x= \frac{\sqrt{3-1}}{2\sqrt{2}}\: and\: y= \frac{1}{\sqrt{2}}$
Hint: Changing $\sin^{-1}x$ to $\cos^{-1}x$ and vice versa; We know that,
$\sin^{-1}x+\cos^{-1}x= \frac{\pi }{2}$
Given: $\cos^{-1}x-\cos^{-1}y$ $= \frac{\pi }{6}$ and $\sin^{-1}x+ \sin^{-1}y = \frac{\pi }{3}$
Here, we have to compute x and y.
Solution:
Let’s replace $\cos^{-1}x-\cos^{-1}y$ by $\sin^{-1}x\: and\: \sin^{-1}y$ , therefore
$\! \! \! \! \! \! \! \! \Rightarrow \cos^{-1}x-\cos^{-1}y = \frac{\pi }{6} \cdot \cdot \cdot (1)\\ \Rightarrow \sin^{-1}x+ \sin^{-1}y = \frac{\pi }{3} \cdot \cdot \cdot \cdot (2)\\ \Rightarrow \frac{\pi }{2} - \sin^{-1}x - \frac{\pi }{2} + \sin^{-1}y = \frac{\pi }{6}\\ -(\sin^{-1}x-\sin^{-1}y) = \frac{\pi }{6} \cdot \cdot \cdot (3)$
Subtracting Equation 3 from 2,
$\! \! \! \! \! \! \! \! \! \sin^{-1}x+\sin^{-1}y +\sin^{-1}x-\sin^{-1}y = \frac{\pi }{3} - \frac{\pi }{6}\\ 2 \sin^{-1}x = \frac{\pi }{6}\\ \sin^{-1}x = \frac{\pi }{12}\\ x = \sin \frac{\pi }{12}\\$
$x= \frac{\sqrt{3}-1}{2\sqrt{2}}$
Adding Equation 2 and 3,
$\! \! \! \! \! \! \! \! \! 2 \sin^{-1}y = \frac{\pi }{3} + \frac{\pi }{6}\\ 2 \sin^{-1}y = \frac{\pi }{2}\\ \sin^{-1}y = \frac{\pi }{4}\\ y = \sin \frac{\pi }{4}\\ y = \frac{1}{\sqrt2}$
Concept: Properties and relations between inverse trigonometric functions.
Note: Values of Trigonometric functions for various radians and degrees.

Inverse Trigonometric Function Exercise 3.10 Question 6

Answer:

Answer: $\frac{1}{5}$
Hint: Convert $\sin^{-1}x$ to $\cos^{-1}x$ or vice versa.
Given: $\sin\left \{\sin^{-1}\frac{1}{5}+ \cos^{-1}x \right \} = 1$
Here, we have to compute x.
Solution:
$\! \! \! \! \! \! \! \! \! \sin^{-1} \frac{1}{5}+ \cos^{-1}x = \sin^{-1}1\\ \Rightarrow \sin^{-1} \frac{1}{5}+ \cos^{-1}x = \frac{\pi }{2}\\ \Rightarrow \cos^{-1}x = \frac{\pi }{2} - \sin^{-1} \frac{1}{5} \; \; \; \; (\because \sin^{-1}x+ \cos^{-1}x= \frac{\pi }{2} )\\ \Rightarrow \cos^{-1}x = \cos^{-1}\frac{1}{5}\\ x = \frac{1}{5}$
Concept: Properties of inverse trigonometric functions.
Note: Deriving value of basic Trigonometric functions or remember all degrees of basic Trigonometric functions.

Inverse Trigonometric Function Exercise 3.10 Question 7

Answer: $\frac{\sqrt{3}}{2}$
Hint: Use $\sin^{-1}x+ \cos^{-1}x= \frac{\pi }{2}$
Given: $\sin^{-1}x=\frac{\pi }{6}+\cos^{-1}x$
Here, we have to compute x.
Solution:
We have, $\sin^{-1}x=\frac{\pi }{6}+\cos^{-1}x$
$\! \! \! \! \! \! \!\! \! \Rightarrow \sin^{-1}x-\cos^{-1}x= \frac{\pi }{6}\\ \Rightarrow \sin^{-1}x-(\frac{\pi }{2}- \sin^{-1}x)= \frac{\pi }{6} \: \: \: \: (\because \sin^{-1}x+ \cos^{-1}x= \frac{\pi }{2})\\ \Rightarrow 2 \sin^{-1}x=\frac{\pi }{6} + \frac{\pi }{2}\\ \Rightarrow 2 \sin^{-1}x = \frac{2\pi }{3}\\ \Rightarrow \sin^{-1}x = \frac{\pi }{3}\\ x = \sin \frac{\pi }{3}\\ x = \frac{\sqrt3}{2}$
Concept: Properties of inverse trigonometric functions.
Note: Remember basic Trigonometric values.

Inverse Trigonometric Function Exercise 3.10 Question 8

Answer: $\frac{1}{2}$
Hint: $\sin^{-1}x+ \cos^{-1}x= \frac{\pi }{2}$
Given: $4 \sin^{-1}x = \pi -\cos^{-1}x$
Here, we have to compute x.
Solution:
We know,
$\sin^{-1}x+ \cos^{-1}x= \frac{\pi }{2}$
Therefore,
$\! \! \! \! \! \! \! \! \! 4 \sin^{-1}x = \pi - (\frac{\pi }{2}-\sin^{-1}x)\\ \Rightarrow 4\sin^{-1} x = \pi - \frac{\pi }{2}+\sin^{-1}x\\ \Rightarrow 4 \sin^{-1}x = \frac{\pi }{2}+\sin^{-1}x\\ \Rightarrow 3 \sin^{-1}x = \frac{\pi }{2}\\ \Rightarrow \sin^{-1}x = \frac{\pi }{6}\\ \Rightarrow x = \sin \frac{\pi }{6}\\ \Rightarrow x = \frac{1}{2}$
Concept: Properties of inverse trigonometric functions.
Note: Remember basic Trigonometric values

Inverse Trigonometric Function Exercise 3.10 Question 9

Answer: $\sqrt{3}$
Hint: Use $\tan^{-1}x + \cot^{-1}x= \frac{\pi }{2}$
Given: $\tan^{-1}x + 2\cot^{-1}x= \frac{2\pi }{3}$
Here, we have to compute x.
Solution:
We have, $\tan^{-1}x + \cot^{-1}x= \frac{\pi }{2}$
Therefore,
$\! \! \! \! \! \! \! \! \Rightarrow \frac{\pi }{2} - \tan^{-1}x = \cot ^{-1}x\\ \Rightarrow \tan^{-1}x + 2 \cot^{-1}x =\frac{2\pi }{3}\\ \Rightarrow \tan^{-1}x + 2 (\frac{\pi }{2} - \tan^{-1}x ) = \frac{2\pi }{3}\\ \Rightarrow \pi - \tan^{-1}x = \frac{2\pi }{3}\\ \Rightarrow \pi - \frac{2\pi }{3} = \tan^{-1}x\\ \Rightarrow 3\pi - \frac{2\pi }{3} = \tan^{-1}x\\ \Rightarrow \frac{\pi }{3} = \tan^{-1}x\\ \Rightarrow x = \tan \frac{\pi }{3}\\\Rightarrow x = \sqrt{3}$
Concept: Properties of inverse trigonometric functions.
Note: Derive value of basic Trigonometric functions.

Inverse Trigonometric Function Exercise 3.10 Question 10

Answer: $1$
Hint: Use $\tan^{-1}x + \cot^{-1}x= \frac{\pi }{2}$
Given: $5\tan^{-1}x + 3\cot^{-1}x= 2\pi$
Here, we have to compute x.
Solution:
As we know, $\tan^{-1}x + \cot^{-1}x= \frac{\pi }{2}$
Therefore,
$\Rightarrow \frac{\pi }{2} - \tan^{-1}x = \cot^{-1}x$
$\Rightarrow 5\tan^{-1}x + 3\cot^{-1}x= 2\pi$
$\Rightarrow 5 \tan^{-1}x + 3 (\frac{\pi }{2} - \tan^{-1}x ) = 2\pi$
$\! \! \! \! \! \! \! \! \! {\Rightarrow }5 \tan^{-1}x - 3 \tan^{-1}x = 2\pi - \frac{3\pi }{2}\\ \Rightarrow 2 \tan^{-1}x = \frac{\pi }{2}\\ \Rightarrow \tan^{-1}x = \frac{\pi }{4}\\ \Rightarrow x = \tan \frac{\pi }{4}\\ \therefore x = 1$

Concept: Properties of inverse trigonometric functions.

Note: Value of basic Trigonometric functions should be memorised.

Students can refer to RD Sharma Class 12 Chapter 3 Exercise 3.10 to learn about inverse trigonometry. The RD Sharma Class 12 Chapter 3 Exercise 3.10 provides detailed step-by-step explanations of textbook problems. Aside from solutions for each chapter of CBSE Class 12 Maths, Career360 also offers Class 12 Maths notes, solved exemplar problems, previous year question papers, and worksheets as study materials for students preparing for their Class 12 board exams.

Mathematics has always been a complex subject for many students. These solutions help students to give their best shot and score well in exams.

These solutions are created by a team of experts with the utmost care. Class 12 RD Sharma Chapter 3 Exercise 3.10 contains answers and hints of a few questions that might be difficult for students. The RD Sharma Class 12 Chapter 3 Exercise 3.10 provides detailed step-by-step explanations of textbook problems. Aside from solutions for each chapter of CBSE Class 12 Maths, Career360 also offers Class 12 Maths notes, solved exemplar problems, previous year question papers, and worksheets as study materials for students preparing for their Class 12 board exams.

RD Sharma Class 12th Exercise 3.10 solution on Career360 provides these solutions free of charge, and every student must use them to benefit. As a result, students have a fantastic opportunity to learn and gain a depth of knowledge from these solutions. Many previous batch students have gained a lot by studying with the RD Sharma solutions books. The students can very well visit the Career 360 website to obtain the access to the best set of solution books.

Chapter-wise RD Sharma Class 12 Solutions

Frequently Asked Questions (FAQs)

1. Why should I refer to the RD Sharma Class 12 Maths Book?

The RD Sharma Class 12 Maths Book is the best book for exam preparation. If you want to take competitive exams like IIT JEE, this book will help you get there. RD Sharma Class 12 Maths Solutions Chapter 3 Exercise 3.5 contains everything you need to know about the subject, including examples, tips, tricks, and practice questions.

2. Are these solutions free and up to date?

Yes, Career360's solutions are free and up to date with the most recent book questions. You can take advantage of the benefits by visiting the website. The high quality of these solutions will make you feel like a knowledge elite.

3. What are the topics covered in Inverse Trigonometry?

The chapter 'Inverse Trigonometry' involves inverses of cosine, sine, tangent, cotangent, secant, and cosecant functions.

4. Is the chapter Inverse Trigonometry hard to learn?

The students are capable of learning any topics with a good amount of practice. Even though Inverse trigonometry is complex, the RD Sharma Class 12 Chapter 3 Exercise 3.10 will lend a helping hand to the students.

5. What makes the RD Sharma solution books stand out from the other books available in the market?

The RD Sharma books have lots of practice problems that gives the students a strong foundation regarding the concepts. Therefore, many people use it making it one of the most trusted books of the class 12 students.