RD Sharma Class 12 Exercise 3.4 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 3.4 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 12:02 PM IST

RD Sharma Class 12 Chapter 3 Exercise 3.4 is the one-stop solution for every student facing problems in learning mathematics, especially inverse trigonometry. Students will be able to solve the exercise 3.4 easily. The concepts and solutions of this chapter are formatted in a way that students will find it easy to solve the questions in multiple ways. RD Sharma class 12th exercise 3.4 solutions are there to help a student to score and compete well in any exam.

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

Inverse Trigonometric Functions Excercise: 3.4

Inverse Trigonometric Function Exercise 3.4 Question 1 (i)

Answer:$\frac{3\pi }{4}$
Hint:$The\: range\: o\! f\: the\: principal\: value \: o\! f \sec ^{-1} is\: \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \}$
Given: $\sec^{-1}\left (- \sqrt{2} \right )$
Solution:$Let ,y= \sec ^{-1}\left ( -\sqrt{2} \right )$
$\sec y= -\sqrt{2}$.
$W\! e\: know\: that \: \sec \frac{\pi }{4}= \sqrt{2}$
$\Rightarrow -\sec \left ( \frac{\pi }{4} \right )= -\sqrt{2}$
$\Rightarrow \sec \left ( \pi -\frac{\pi }{4} \right )$
$\Rightarrow \sec \left ( \frac{3\pi }{4} \right )$
$Thus, the\: range\: o\! f \: the\: principal\: value\: o\! f \: \sec ^{-1} is \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \}$
$\Rightarrow \sec \left ( \frac{3\pi }{4} \right )= -\sqrt{2}$$Hence, the\: principal \: value\: o\! f \sec ^{-1} \left ( \sqrt{2} \right )is\frac{3\pi }{4} .$

#### Inverse Trigonometric Function Exercise 3.4 Question 1 (ii)

Answer: $\frac{\pi }{3}$
Hint: The range of principal value of $\sec ^{-1} is \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \}$
Given:$\sec ^{-1}\left ( 2 \right )$
Solution:$Let \: y= \sec ^{-1}\left ( 2 \right )$
$\sec y= 2$
As we know$\sec\left ( \frac{\pi }{3} \right ) = 2$
Therefore the range of principal value of $\sec ^{-1} \: is \: \left [ 0,\pi \right ]\! -\! \left \{ \frac{\pi }{2} \right \}\! and\, \sec \left ( \frac{\pi }{3} \right )= 2$
Thus the principal value of $\sec ^{-1}\left ( 2 \right )\: is \frac{\pi }{3}$

Inverse Trigonometric Function Exercise 3.4 Question 1 (iii)

Answer:$\frac{\pi }{4}$
Hint: The range of principal value of $\sec ^{-1} is \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \}$
Given: $\sec ^{-1}\left ( 2\sin \left ( \frac{3\pi }{4} \right ) \right )$
Solution: We know that $\sin \left ( \frac{3\pi }{4} \right )= \frac{1}{\sqrt{2}}$
$\therefore 2\sin \left ( \frac{3\pi }{4} \right )\Rightarrow 2\times \frac{1}{\sqrt{2}}$
$\therefore 2\sin \left ( \frac{3\pi }{4} \right )\Rightarrow \sqrt{2}\; \; \; \; \;\; \; \; \; \; \left [ \sqrt{2}\times \sqrt{2}= 2\right ]$
By substituting these values in$\sec ^{-1}\left ( 2\sin \left ( \frac{3\pi }{4} \right ) \right ), we\: get$
$\sec ^{-1}\left ( \sqrt{2} \right )$
$Let\: y= \sqrt{2}$
$\sec \left ( \frac{\pi }{4} \right )= \sqrt{2}$
Therefore the range of principal value of $\sec ^{-1}is\left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \} and \sec \left ( \frac{\pi }{4} \right )\! =\! \sqrt{2}$
Thus,the principal value of $\sec ^{-1}\left ( 2\sin \left ( \frac{3\pi }{4} \right ) \right ) is .\: \frac{\pi }{4}$

Inverse Trigonometric Function Exercise 3.4 Question 1 (iv)

Answer:$\frac{2\pi }{3}$
Hint: The range of the principal value of $\sec ^{-1} is \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \}$
Given: $\sec ^{-1}\left ( 2\tan \left ( \frac{3\pi }{4} \right ) \right )$
Solution: We know that $\tan \left ( \frac{3\pi }{4} \right )= -1$
$\therefore 2\tan \left ( \frac{3\pi }{4} \right )\Rightarrow 2\times -1$
$2\tan \left ( \frac{3\pi }{4} \right )\Rightarrow -2$
By substituting these values in$\sec ^{-1}\left ( 2\tan \left ( \frac{3\pi }{4} \right ) \right )$,we get
$\sec ^{-1}\left ( -2 \right )$
Let ,$y= \sec ^{-1}\left ( -2 \right )$
$\sec \left ( y \right )= -2$
$-\sec\left ( \frac{\pi }{3} \right )= -2$
$\Rightarrow \sec\left ( \pi - \frac{\pi }{3} \right )$
$\Rightarrow \sec\left ( \frac{2\pi }{3} \right )$
$\therefore$The range of the principal value of $\sec ^{-1} is \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \} and\, \sec \left ( \frac{2\pi }{3} \right )\! =\! -2$
$\therefore$The principal value of $\sec ^{-1}\left ( 2\tan \left ( \frac{3\pi }{4} \right ) \right )\, is\, \frac{2\pi }{3}$

Inverse Trigonometric Function Exercise 3.4 Question 2 (i)

Answer:$-\frac{\pi }{3}$
Hint: The range of the principal value of $\sec ^{-1} is \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \}$
The range of the principal value of $\tan^{-1} is \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ]$
Given:$\tan^{-1}\left ( \sqrt{3} \right )-\sec ^{-1}\left ( -2 \right )$
Solution:First we solve $\tan^{-1}\left ( \sqrt{3} \right )$
Let,$y= \tan^{-1}\left ( \sqrt{3} \right )$
$\tan y=\left ( \sqrt{3} \right )$
$\tan y=\tan \left ( \frac{\pi }{3} \right )$
Since$\left [ \tan \left ( \frac{\pi }{3} \right )= \sqrt{3} \right ]$
$\therefore y= \frac{\pi }{3}$
Since range of $\tan^{-1} is \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ]$

Hence, Principal value is $\frac{\pi }{3}$

Now, Solving $\sec ^{-1}\left ( -2 \right )$

Let,$y= \sec ^{-1}\left ( -2 \right )$

We know that

$\sec ^{-1}\left ( -x \right )= \pi -\sec ^{-1}\left ( x \right )$
$y=\pi-\sec ^{-1}(2)$ $\left\{\begin{array}{l} \because \sec \frac{\pi}{3}=2 \\ \frac{\pi}{3}=\sec ^{-1} 2 \end{array}\right\}$

$y= \pi -\frac{\pi }{3}$
$y=\frac{2\pi }{3}$
Since range of the principal value of $\sec ^{-1} is \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \}$
Hence, principal value is $\frac{2\pi }{3}$
Now, we have
$\tan ^{-1} \left ( \sqrt{3} \right )= \frac{\pi }{3}\, and \, \sec ^{-1} \left ( -2 \right )= \frac{2\pi }{3}$
Solving
$\tan ^{-1}\left ( \sqrt{3} \right )-\sec ^{-1}\left ( -2 \right )$
$\Rightarrow \frac{\pi }{3}-\frac{2\pi }{3}$
$\Rightarrow \frac{\pi-2\pi }{3}$
$\Rightarrow- \frac{\pi}{3}$

Inverse Trigonometric Function Exercise 3.4 Question 2 (ii)

Answer:$\frac{-2\pi }{3}$
Hint:The range of the principal value of $\sec ^{-1} is \left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \}$
The range of the principal value of $\sin ^{-1} is \left [ \frac{-\pi }{2} ,\frac{\pi }{2}\right ]$
Given: $\sin ^{-1}\left ( \frac{\sqrt{3}}{2} \right )-2\sec ^{-1}\left ( 2\tan \frac{\pi }{6} \right )$
Solution: First we solve $\sin ^{-1}\left ( \frac{\sqrt{3}}{2} \right )$
Let $\, y= \sin ^{-1}\left ( \frac{\sqrt{3}}{2} \right )$
$\Rightarrow \sin y= \left ( \frac{\sqrt{3}}{2} \right )$
$\Rightarrow -\sin \left ( \frac{\pi }{3} \right )$
$\Rightarrow$As we know $\sin \left ( -\theta \right )= -\sin \left ( \theta \right )$
$-\sin \left ( \frac{\pi }{3} \right )= \sin \left (- \frac{\pi }{3} \right )$
The range of the principal value of $\sin ^{-1} is\left [ \frac{-\pi }{2},\frac{\pi }{2} \right ] and \: \sin \left ( -\frac{\pi }{3} \right )= -\frac{\sqrt{3}}{2}$
Hence, Principal value is $-\frac{\pi }{3}$
Now, we solve $2\tan \left ( \frac{\pi }{6} \right )$
Let us assume $2\tan \left ( \frac{\pi }{6} \right )= \theta$
We know that $\tan \left ( \frac{\pi }{6} \right )= \frac{1}{\sqrt{3}}$
$2 \tan \left ( \frac{\pi }{6} \right )= 2\times \frac{1}{\sqrt{3}}$
$2 \tan \left ( \frac{\pi }{6} \right )= \frac{2}{\sqrt{3}}$
Now by substituting these values in $\sec ^{-1}\left ( 2\tan \frac{\pi }{6} \right )$ we get
$\sec ^{-1}\left ( \frac{2}{\sqrt{3}} \right )$
Now let $\sec ^{-1}\left ( \frac{2}{\sqrt{3}} \right )= z$
$\sec z= \frac{2}{\sqrt{3}}$
$\sec \left ( \frac{\pi }{6} \right )= \frac{2}{\sqrt{3}}$
The range of the principal value of $\sec ^{-1} is\left [ 0,\pi \right ]-\left \{ \frac{\pi }{2} \right \} and\: \sec \left ( \frac{\pi }{3} \right )= \frac{2}{\sqrt{3}}$
Hence, Principal value of $\sec ^{-1}\left ( 2\tan \left ( \frac{\pi }{6} \right ) \right )\, is\, \frac{\pi }{3}$
Now, We have
$\sec ^{-1}\left ( 2\tan \left ( \frac{\pi }{6} \right ) \right )= \frac{\pi }{3} and\: \sin ^{-1}\left ( \frac{\sqrt{3}}{2} \right ) = -\frac{\pi }{3}$
Solving
$\sin ^{-1}\left ( \frac{\sqrt{3}}{2} \right )-2\sec ^{-1}\left ( 2\tan \left ( \frac{\pi }{6} \right ) \right )$
$\Rightarrow -\frac{\pi }{3}-\frac{\pi }{3}$
$\Rightarrow \frac{-2\pi }{3}$

Inverse Trigonometric Function Exercise 3.4 Question 3 (i)

Answer:$\left (-\infty,0\right ] \cup \left [ \frac{2}{3},\infty \right )$
Hint:$Domain\left ( y^{-1} \right )= Range\left ( y \right )$
$Domain\, o\! f\sec ^{-1}\left ( x \right )=\left ( -\infty,-1 \right ]or\left [ 1,\infty \right ]$
Given: $\sec ^{-1}\left ( 3x-1 \right )$
Solution:$W\! e\, know \, that\, the\, range\, o\! f\sec x\, is \left ( -\infty,-1 \right )\cup \left ( 1,\infty \right )$
$3x-1< -1\: and \: 3x-1\geq 1$
$3x< -1+1\: and \: 3x\geq 1+1$
$3x< 0 \: and\: 3x\geq 2$
$x< 0\: and\: x\geq \frac{2}{3}$
$x\: \epsilon \: \left ( -\infty,0 \right ]and\: x\: \epsilon \left [ \frac{2}{3},\infty \right ]$
$x\: \epsilon \: \left ( -\infty,0 \right ]\cup \left [ \frac{2}{3},\infty \right ]$
$Domain\, is \left (-\infty,0\right ] \cup \left [ \frac{2}{3},\infty \right ]$

Inverse Trigonometric Function Exercise 3.4 Question 3 (ii)

Answer:$\left ( -\infty,-1 \right ] \cup \left [ 1,\infty\right )$
Hint: Domain of $\sec ^{-1}\left ( x \right ) is \:\left ( -\infty,-1\right ] \cup \left [ 1,\infty \right )$
Domain of $\tan ^{-1}\left ( x \right )$ is $R$
Given: $\sec ^{-1}\left ( x \right )-\tan ^{-1}x$
Solution:We know that the domain of $\sec ^{-1}\left ( x \right ) is \left ( -\infty,1 \right ]\cup \left [ 1,\infty \right ) \cdot \cdot \cdot \cdot \cdot (i)$
Domain of $\tan ^{-1}x\: is \: R \cdot \cdot \cdot \cdot \cdot \cdot (ii)$
Union of (i) and (ii) will be domain of the given function
$\left ( -\infty,-1 \right ] \cup \left [ 1,\infty\right )\cup R$
$\Rightarrow \left ( -\infty,-1 \right ] \cup \left [ 1,\infty\right )$
The domain of the given function is $\left ( -\infty,-1 \right ] \cup \left [ 1,\infty\right ).$

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Chapter-wise RD Sharma Class 12 Solutions

1. What  are the six trigonometric functions?

Sine, Cosine, Tangent, Secant, Cosecant, and Cotangent are the six trigonometric functions

2. What are the three most fundamental trigonometric functions?

Sine, Cosine, and Tangent are the three fundamental trigonometric functions.

3. What is the value of sin θ, cos θ and tan θ, if θ=30 degrees?

If θ = 30 degrees, then,

Sin θ = sin 30 = ½

Cos θ = cos 30 = √3/2

Tan θ = tan 30 = 1/√3

4. Are the solutions provided in the RD Sharma class 12 books prepared from the exam point of view?

Yes, the solutions provided in the RD Sharma class 12 books are prepared from the exam point of view as these solutions are prepared by experts and are constructed from the exam point of view.

5. Is the RD Sharma Chapter 3 Solutions book enough for the CBSE board students to prepare for their public exams?

The RD Sharma Solutions books are more than enough for the class 12 students to prepare for their board exams. In addition, these books provide a deeper insight into the concepts to the students hence making it easier for them to learn.

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