RD Sharma Solutions Class 12 Mathematics Chapter 3 VSA

Q: Does this material have all chapters?

Yes, Career360 offers all the chapters on their website, available for free and accessible by students. You can learn from these solutions to score good marks in exams. For example, to get more insight on Inverse Trigonometry, you can download Class 12 RD Sharma Chapter 3 VSAQ Solution.

Q: What are the applications of trigonometry?

These are some of the applications of trigonometry: Oceanography - To calculate tide heights. Calculus Physics - Sine and Cos functions are used as wave representations. Height calculation - To Measure the height of buildings or mountains.

Q: Who invented Inverse Trigonometry?

Daniel Bernoulli introduced inverse trigonometry. This concept was used to determine the angle if two sides of a triangle are given. For more information on Inverse Trigonometry, check out RD Sharma Class 12 Solutions Inverse Trigonometric Functions VSAQ

Q: Are NCERT and RD Sharma books similar?

NCERT books for maths are good for general preparation, but RD Sharma books are relatively better to dive deep into the concepts. They are widely considered by teachers and students all over the country and are excellent means of studying for exams. To find the RD Sharma material for Inverse Trigonometry, you can study from RD Sharma Class 12 Solutions Chapter 3 VSAQ.

Q: Where is inverse trigonometry used in real life?

nverse trigonometry is a concept through which we can determine the third side of a triangle if two are known. This has its applications across engineering physics, geometry, and navigation on land and sea. To get more information about Inverse Trigonometry, you can refer to RD Sharma Class 12 Solutions Inverse Trigonometric Functions VSAQ.

RD Sharma Solutions Class 12 Mathematics Chapter 3 VSA

Updated on Jan 21, 2022 12:37 PM IST

RD Sharma has been the choice of many students and teachers all over the country. Their simple easy to understand solutions make it very beneficial for students to learn the concepts quickly. Many teachers use these materials for lectures and to set up question papers. As these contain numerous examples, students can be thorough with their preparation and leave no problem unsolved.

Also Read - RD Sharma Solution for Class 9 to 12 Maths

Inverse Trigonometric Functions Excercise: VSAQ

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

Inverse Trigonometric Function Exercise Very Short Answer Question 1

Answer:
$C=\frac{a+b}{2}$
Hint:
Find differentiation of $f\left ( x \right )$ and apply Rolle’s Theorem.
Given:
$f\left ( x \right )=Ax^{2}+Bx+C$ and $f\left ( a \right )=f\left ( b \right )$
Solution:
$f\left ( a \right )=f\left ( b \right )$
$\begin{array}{ll} \Rightarrow & A a^{2}+B a+C=A b^{2}+B b+C \\\\ \Rightarrow & A a^{2}+B a=A b^{2}+B b \\\\ \Rightarrow & A\left(a^{2}-b^{2}\right)+B(a-b)=0 \\\\ \Rightarrow & A(a-b)(a+b)+B(a-b)=0 \\\\ \Rightarrow & (a-b)[A(a+b)+B]=0 \\\\ \Rightarrow & A(a+b)+B=0 \text { and } a-b=0 \end{array}$

$\Rightarrow A=\frac{-B}{a+b}$ and $a-b$
$\Rightarrow A=\frac{-B}{a+b}$ $\left [ \because a\neq b \right ]$
Now, $f\left ( x \right )=Ax^{2}+Bx+C$
${f}'\left ( x \right )=2xA+B$
${f}'\left ( C \right )=2AC+B$
By Rolle’s Theorem,
$\Rightarrow$ ${f}'\left ( C \right )=0$
$\Rightarrow$ $2AC+B=0$
$\Rightarrow$ $C=\frac{-B}{2A}$
$\Rightarrow$ $C=\frac{-B}{2\left ( \frac{-B}{a+b}\right )}$ $\left [ \because A=\frac{-B}{a+b} \right ]$
$\Rightarrow$ $C=\frac{-B}{2}\times\frac{a+b}{\left ( -B \right )}$
$\Rightarrow$ $C=\frac{a+b}{2}$

Inverse Trigonometric Function Exercise Very Short Answer Question 2

Answer:
Statement of Rolle’s Theorem
Hint:
You must know about Rolle’s Theorem.
Given:
Rolle’s Theorem
Solution:
Rolle’s Theorem states that if $f\left ( x \right )$ is continuous in the closed interval $\left [ a,b \right ]$ and differentiable on open interval $\left ( a,b \right )$ such that $f\left ( a \right )=f\left ( b \right )$ then ${f}'\left ( x \right )=0$ for some $a< x< b$.

Inverse Trigonometric Function Exercise Very Short Answer Question

Answer:
Statement of mean value theorem.
Hint:
You must know about mean value theorem.
Given:
Lagrange’s mean value theorem.
Solution:
Let be a function that satisfies the following hypothesis.

$f\left ( x \right )$ is continuous on the closed interval $\left [ a,b \right ]$.
$f\left ( x \right )$ is differentiable on the open interval $\left ( a,b \right )$.

Then there is a value in $\left ( a,b \right )$ such that
${f}'\left ( c \right )=\frac{f\left ( b \right )-f\left ( a \right )}{b-a}$
Or equivalently
$f\left ( b \right )-f\left ( a \right )={f}'\left ( c \right )\left ( b-a \right )$

Inverse Trigonometric Function Exercise Very Short Answer Question 4

Answer:
$n=3$
Hint:
Find differentiate of $f\left ( x \right )$ and then apply Rolle’s Theorem.
Given:
$f\left ( x \right )=2x\left ( x-3 \right )^{n},x\: \epsilon \left [ 0,2\sqrt{3} \right ]$and value of $c$ is $\frac{3}{4}$.
Solution:
According to Rolle’s Theorem
$\begin{array}{ll} & f(a)=f(b) \\\\ \Rightarrow & f(0)=f(2 \sqrt{3}) \\\\ \Rightarrow & 0=2 \cdot 2 \sqrt{3}(2 \sqrt{3}-3)^{n} \end{array}$
$\Rightarrow \left ( 2\sqrt{3}-3 \right )^{n}=0$
Now, $f\left ( x \right )=2x\left ( x-3 \right )^{n}$
Differentiate with respect to $x$,
${f}'\left ( x \right )=2\left ( x-3 \right )^{n}+2x\cdot n\left ( x-3 \right )^{n-1}$ $\left [ \because \frac{d\left ( uv \right )}{dx}=u\frac{dv}{dx} +v\frac{du}{dx}\right ]$
Applying Rolle’s Theorem,
${f}'\left ( c \right )=2\left ( c-3 \right )^{n}+2c\cdot n\left ( c-3 \right )^{n-1}$
${f}'\left ( c \right )=0$
$2\left ( c-3 \right )^{n}+2c\cdot n\left ( c-3 \right )^{n-1}=0$
$\Rightarrow 2\left ( \frac{3}{4}-3 \right )^{n}+2\cdot \frac{3}{4}\cdot n\left ( \frac{3}{4}-3 \right )^{n-1}=0$ $\left [ \because c=\frac{3}{4} \right ]$
$\Rightarrow \left ( \frac{3}{4}-3 \right )^{n}\left [ 2\left ( \frac{3}{4}-3 \right )+\frac{3n}{2}\right ]=0$
$\Rightarrow \left ( \frac{3}{4}-3 \right )^{n-1}=0$ or $2\left ( \frac{3-12}{4} \right )+\frac{3n}{2}=0$
$\Rightarrow$ $\frac{3-12+3n}{2}=0$
$3-12+3n=0$
$3n=9$
$n=3$

Inverse Trigonometric Function Exercise Very Short Answer Question 5

Answer:
$c=\sqrt{5}$
Hint:
Find differentiability of $f\left ( x \right )$ and then apply the formula of ${f}'\left ( c \right )$.
Given:
$f\left ( x \right )=\sqrt{x^{2}-4},x\: \epsilon \left [ 2,3 \right ]$
Solution:
$f\left ( x \right )=\sqrt{x^{2}-4}$
$f\left ( x \right )=\left ( x^{2}-4 \right )^{\frac{1}{2}}$
Differentiate,
$\Rightarrow {f}'\left ( x \right )=\frac{1}{2}\left ( x^{2}-4 \right )^{\frac{1}{2}-1}\cdot 2x$ $\left [ \because \frac{d}{dx}\left ( x^{n} \right ) =n\left ( x \right )^{n-1}\right ]$
$\begin{aligned} \Rightarrow &f^{\prime}(x)=\frac{1}{2}\left(x^{2}-4\right)^{\frac{-1}{2}} \cdot 2 x \\\\ \Rightarrow &f^{\prime}(x)=\frac{1}{2 \sqrt{x^{2}-4}} \cdot 2 x \\\\ \Rightarrow &f^{\prime}(x)=\frac{x}{\sqrt{x^{2}-4}} \\\\ \Rightarrow &f^{\prime}(c)=\frac{c}{\sqrt{c^{2}-4}} \end{aligned}$
Use mean value theorem,
$\begin{aligned} & f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\\\ \Rightarrow & \frac{c}{\sqrt{c^{2}-4}}=\frac{\sqrt{5}-0}{3-2} \\\\ \Rightarrow & \frac{c}{\sqrt{c^{2}-4}}=\sqrt{5} \\\\ \Rightarrow & c=\sqrt{5}\left(\sqrt{c^{2}-4}\right) \\\\ \Rightarrow & c^{2}=5\left(c^{2}-4\right) \end{aligned}$
$\Rightarrow c^{2}=5c^{2}-20$
$\Rightarrow4 c^{2}=20$
$\Rightarrow c^{2}=5$
$\Rightarrow c=\pm \sqrt{5}$ $\left [ \because x\: \epsilon \left [ 2,3 \right ] \right ]$
$\Rightarrow c=\sqrt{5}$

Inverse Trigonometric Functions exercise Very short answer question 1

Answer: $\frac{\pi}{3}$
Given: $\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)+\cos ^{-1}\left(\frac{-1}{2}\right)$
Hint: $\sin ^{-1}(-x)=-\sin ^{-1} x, x \in(-1,1)$
$\cos ^{-1}(-x)=\pi-\cos ^{-1} x, x \in(-1,1)$
Solution:
$\begin{array}{l} \sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)+\cos ^{-1}\left(\frac{-1}{2}\right)=-\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\pi-\cos ^{-1}\left(\frac{1}{2}\right) \\\\ =\sin ^{-1}\left(\sin \frac{\pi}{3}\right)+\pi-\cos ^{-1}\left(\cos \frac{\pi}{3}\right) \\\\ =\frac{-\pi}{3}+\pi-\frac{\pi}{3} \\\\ =\frac{\pi}{3} \end{array}$

Inverse Trigonometric Functions exercise Very short answer question 2

Answer: $\pi$
Given: $\sin ^{-1} x, x \in(-1,1)$
Hint: The maximum value of $\sin ^{-1} x$ in $x \in (-1,1)$ is at 1 .
Solution:
$\sin ^{-1}(1)$ maximum value is,
$=\sin ^{-1}\left(\sin \frac{\pi}{2}\right)$
$=\frac{\pi}{2}$
Again the minimum value is at -1
So,
$\begin{aligned} \sin ^{-1}(-1) &=-\sin ^{-1}(1) \\\\ &=-\sin ^{-1}\left(\frac{\pi}{2}\right) \\\\ &=-\frac{\pi}{2} \end{aligned}$
The difference between the minimum value, maximum value is
$\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right)=\pi$

Inverse Trigonometric Functions exercise Very short answer question

Answer:
Given: $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2}, x+y+z=?$
Hint: The maximum value in the range of $\sin ^{-1}$ is $\frac{\pi}{2}$
Solution:
$\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{\pi}{2}+\frac{\pi}{2}+\frac{\pi}{2}$
Here sum of three inverse of $\sin x$ is terms $\frac{\pi}{2}$, i.e. every $\sin$ inverse function is equal to $\frac{\pi}{2}$.
$\begin{array}{l} \sin ^{-1} x=\frac{\pi}{2}, \sin ^{-1} y=\frac{\pi}{2}, \sin ^{-1} z=\frac{\pi}{2} \\\\ x=\frac{\sin \pi}{2}, y=\sin \frac{\pi}{2}, z=\sin \frac{\pi}{2} \\\\ x=1, y=1, z=1 \\\\ x+y+z=1+1+1 \\\\ \quad=3 \end{array}$

Inverse Trigonometric Functions exercise Very short answer question 4

Answer:$\pi-2 \tan ^{-1} x$
Given:
$x>1, \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
Hint:
$2 \tan ^{-1} x=\pi-\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) for x>1$
Solution:
$\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=\pi-2 \tan ^{-1} x$

Inverse Trigonometric Functions exercise Very short answer question 5

Answer: $-2 \tan ^{-1} x$
Given:
$x<0, \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$
Hint:
$\frac{1-\tan ^{2} x}{1+\tan ^{2} x}=\cos 2 x$
Solution:
Let, $x=\tan y\\\\$
$\begin{aligned} \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) &=\cos ^{-1}\left(\frac{1-\tan ^{2} y}{1+\tan ^{2} y}\right) \\ &=\cos ^{-1}(\cos 2 y) \\ &=2 y-(1) \end{aligned}$
The value of x is negative, so let x=-a where a>0 .
$\begin{array}{l} -a=\tan y \\\\ y=\tan ^{-1}(-a) \end{array}$

Now,
$\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 y$ Using (1)
$\begin{array}{l} =2 \tan ^{-1}(-a) \\\\ =2 \tan ^{1} x \quad[\therefore x=-a] \end{array}$

Inverse Trigonometric Functions exercise Very short answer question 6

Answer: $\frac{\pi}{2}$
Given:
$\tan ^{-1} x+\tan ^{-1}\left(\frac{1}{x}\right) \ \ \text \ \ { for } \ \ x>0$
Hint:
$\tan ^{-1} x+\tan ^{-1}\left(\frac{1}{x}\right)=\tan ^{-1}\left(\frac{x+\frac{1}{x}}{1-x\frac{1}{x}}\right), x>0$
Solution:
$\begin{aligned} \tan ^{-1} x+\tan ^{-1}\left(\frac{1}{x}\right) &=\tan ^{-1}\left(\frac{x^{2}+1}{0}\right) \\ &=\tan ^{-1}(\infty) \\ &=\tan ^{-1}\left(\tan \frac{\pi}{2}\right) \\ &=\frac{\pi}{2} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 7

Answer:$-\frac{\pi}{2}$
Given:
$\tan ^{-1} x+\tan ^{-1}\left(\frac{1}{x}\right) \text { for } x<0$
Hint:
$\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ when $x<0, \frac{1}{x}<0$
Solution:
$Let \ \ x=-y, y>0\$
Then,
$\begin{aligned} \tan ^{-1} x+\tan ^{-1}\left(\frac{1}{x}\right) &=\tan ^{-1}(-y)+\tan ^{-1}\left(\frac{-1}{y}\right) \\ &=-\left(\tan ^{-1} y+\tan ^{-1} \frac{1}{y}\right) \\ &=-\tan ^{-1}\left(\frac{y+\frac{1}{y}}{1-y_{y}^{1}}\right), y>0 \\ &=-\tan ^{-1}\left(\frac{y^{2}+1}{0}\right) \\ &=-\tan ^{-1}(\infty) \\ &=-\tan ^{-1}\left(\tan \frac{\pi}{2}\right)=-\frac{\pi}{2} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 8

Answer: $\pi$
Given:
$\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$
Hint:
Range of $\sin \left(\frac{-\pi}{2}, \frac{\pi}{2}\right) ; \frac{\pi}{3} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$
Range of $\cos [0, \pi], \frac{2 \pi}{3} \in(0, \pi)$
Solution:
$\begin{aligned} \cos ^{-1}\left(\cos \frac{2 x}{3}\right)+\sin ^{-1}\left(\sin \frac{2 x}{3}\right) &=\cos ^{-1}\left(\cos \frac{2 x}{3}\right)+\sin ^{-1}\left[\sin \left(\frac{2 x}{3}\right)\right] \\ &=\frac{2 \pi}{3}+\pi-\frac{2 \pi}{3} \\ &=\pi \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 9

Answer: 0
Given:
$-1<x<0, \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$
Hint: Try to convert the $\sin$ and $\cos$ function into $\tan$ .
Solution:
$\text { Let, } x=-\tan y, 0<y<\frac{\pi}{2} \\$
$\begin{aligned} \quad \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) &=\sin ^{-1}\left(-\frac{2 \tan y}{1+\tan ^{2} y}\right)+\cos ^{-1}\left(\frac{1-\tan ^{2} y}{1+\tan ^{2} y}\right) \\ &=\sin ^{-1}\{-\sin (2 y)\}+\cos ^{-1}\{\cos (2 y)\} \\ &=-\sin ^{-1}\{\sin 2 y\}+\cos ^{-1}\{\cos (2 y)\} \\ &=-2 y+2 y \\ &=0 \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 10

Answer: $\frac{1}{\sqrt{x^{2}+1}}$
Given: $\sin \left(\cot ^{-1} x\right)$
Hint: $\sin \left(\cot ^{-1} x\right)=\sin \left(\tan ^{-1} \frac{1}{x}\right)\\\\$
$\tan ^{-1} x=\sin \left(\frac{x}{\sqrt{1+x^{2}}}\right)$
Solution:
$\cot ^{-1} x=\tan ^{-1} \frac{1}{x}$
$\sin \left(\cot ^{-1} x\right) =\sin \left(\tan ^{-1} \frac{1}{x}\right) \\$
$=\sin \left[\sin ^{-1}\left(\frac{\frac{1}{x}}{\sqrt{x^{2}+\frac{1}{x}}}\right)\right] \\$
$=\sin \left(\sin ^{-1} \frac{1}{\sqrt{x^{2}+1}}\right) \\\\$
$=\frac{1}{\sqrt{x^{2}+1}} \quad \therefore\left(\sin \left(\sin ^{-1} x\right)=x\right)$

Inverse Trigonometric Functions exercise Very short answer question 11

Answer: $\frac{2 \pi}{3}$
Given:
$\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)$
Hint: The range of $\sin is \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right), \ \ range \ \ of \ \ \cos is \ \ \((0, \pi)$
Solution:

$\begin{aligned} \cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right) &=\cos ^{-1}\left(\cos \frac{\pi}{3}\right)+2 \sin ^{-1}\left(\sin \frac{\pi}{6}\right) \\ \\&=\frac{\pi}{3}+2\left(\frac{\pi}{6}\right) \\\\ &=\frac{\pi}{3}+\frac{\pi}{3} \\ \\&=\frac{2 \pi}{3} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 12

Answer:$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
Given:
Range of $\tan ^{-1} x$
Hint: Range of $\tan$ function should be all real number.
Solution:
$\tan ^{-1} x=\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Inverse Trigonometric Functions exercise Very short answer question 1

Answer: $100 ^\circ$
Given:
$\cos ^{-1}\left(\cos 1540^{\circ}\right)$
Hint: Try to separate the degree into two parts.
Solution:
$\begin{aligned} \cos ^{-1}\left(\cos 1540^{\circ}\right) &=\cos ^{-1}\left\{\cos \left(1440^{\circ}+100^{\circ}\right)\right\} \\ &=\cos ^{-1}\left\{\cos \left(100^{\circ}\right)\right\} \quad \therefore\left[\cos \left(4 \pi+100^{\circ}\right)=\cos 100^{\circ}\right] \\ &=100^{\circ} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 14

Answer: $60^{\circ}$
Given:
$\sin ^{-1}\left(\sin \left(-600^{\circ}\right)\right)$
Hint:
$\sin x=\sin (\pi-x)$
Solution:
$\begin{array}{l} \sin ^{-1}\{\sin x\}=x \\ \end{array}$

$\begin{aligned} \begin{array}{l} \sin ^{-1}\left\{\sin \left(600^{\circ}\right)\right\} =\sin ^{-1}\left\{\sin \left(720^{\circ}-600^{\circ}\right)\right\} \\ \end{array} \end{aligned}$
$\begin{aligned} \begin{array}{l} =\sin ^{-1}\left\{\sin \left(120^{\circ}\right)\right\} \\ =\sin ^{-1}\left\{\sin \left(180^{\circ}-120^{\circ}\right)\right\} \\ =\sin ^{-1}\left\{\sin \left(60^{\circ}\right)\right\}=60^{\circ}\\ \ \\ \end{array} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 15

Answer: $\frac{7}{9}$
Given:
$\cos \left(2 \sin ^{-1} \frac{1}{3}\right)$
Hint: Try to solve the bracket portion
i.e. $\sin ^{-1}$ function.
Solution:
$Let\ \ , y=\sin ^{-1} \frac{1}{3} \\\\ Then\ \ , \sin y=\frac{1}{3}$
Now,
$\begin{array}{l} \cos y=\sqrt{1-\sin ^{2} y}, \cos y=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3} \\\\ \cos \left(2 \sin ^{-1} \frac{1}{3}\right)=\cos 2 y \\\\ \end{array}$
$\begin{array}{l} =\cos ^{2} y-\sin ^{2} y \quad\left[\therefore \cos 2 x=\cos ^{2} x-\sin ^{2} x\right] \\\\ =\left(\frac{2 \sqrt{2}}{3}\right)^{2}-\left(\frac{1}{3}\right)^{2} \\\\ =\frac{8}{9}-\frac{1}{9} \\\\ =\frac{7}{9} \end{array}$

Inverse Trigonometric Functions exercise Very short answer question 16

Answer: $70 ^\circ$
Given:
$\sin ^{-1}\left(\sin 1550^{\circ}\right)$
Hint:
$\sin ^{-1}(\sin x)=x$
Solution:
We, know that
$\sin ^{-1}(\sin x)=x$
Now,
$\begin{aligned} \sin ^{-1}\left(\sin 1550^{\circ}\right) &=\sin ^{-1}\left\{\sin \left(1620^{\circ}-1550^{\circ}\right)\right\} \\ &=\sin ^{-1}\left(\sin 70^{\circ}\right) \\ &=70^{\circ} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 17

Answer: $\frac{1}{\sqrt{10}}$
Given:
$\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)$
Hint:
$\sin ^{-1}(\sin x)=x$
Solution:
$\begin{array}{l} \cos ^{-1} x=2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}} \\ \tan ^{-1} x=\sin ^{-1} \frac{x}{\sqrt{1+x^{2}}} \\ \begin{aligned} \therefore \sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right) &=\sin \left(\frac{1}{2}\left(2 \tan ^{-1} \sqrt{\frac{1-\frac{4}{5}}{1+\frac{4}{5}}}\right)\right) \\ &=\sin \left(\tan ^{-1} \sqrt{\frac{\frac{1}{5}}{\frac{9}{5}}}\right) \\ &=\sin \left(\tan ^{-1} \frac{1}{3}\right) \\ & \end{aligned} \\ \end{array}$
$\begin{array}{l} =\sin \left\{\sin ^{-1}\left(\frac{\frac{1}{2}}{\sqrt{1+\frac{1}{9}}}\right)\right\} \\\\ =\sin \left\{\sin ^{-1} \frac{1}{\sqrt{10}}\right\} \\\\ =\frac{1}{\sqrt{10}} \end{array}$

Inverse Trigonometric Functions exercise Very short answer question 18

Answer: $\frac{3}{5}$
Given:
$\sin \left(\tan ^{-1} \frac{3}{4}\right)$
Hint: Try to solve $\tan$ function first.
Solution:
$\begin{array}{l} \tan ^{-1} x=\sin ^{-1} \frac{x}{\sqrt{1+x^{2}}} \\\\\end{array}$
$\begin{array}{l} \sin \left(\tan ^{-1} \frac{3}{4}\right)=\sin \left\{\sin ^{-1}\left(\frac{\frac{3}{4}}{\sqrt{1+\frac{9}{16}}}\right)\right\} \\\\ \\\\\end{array}$
$\begin{array}{l} \qquad=\sin \left\{\sin ^{-1}\left(\frac{\frac{2}{4}}{\frac{5}{4}}\right)\right\} \\\\ \\\\\end{array}$
$\begin{array}{l} \\\\ =\sin \left\{\sin ^{-1} \frac{3}{5}\right\} \\\\ =\frac{3}{5}\\\\\end{array}$

Inverse Trigonometric Functions exercise Very short answer question 19

Answer: $\pi$
Given:
$\cos ^{-1}\left(\tan \frac{3 \pi}{4}\right)$
Hint: Try to solve $\tan$ function first.
$\therefore \tan (\pi-x)=\tan x$
Solution:
$\begin{aligned} \cos ^{-1}\left(\tan \frac{3 \pi}{4}\right) &=\cos ^{-1}\left\{-\tan \left(\pi-\frac{3 \pi}{4}\right)\right\} \\\\ &=\cos ^{-1}\left\{\tan \left(\frac{-\pi}{4}\right)\right\} \\\\ &=\cos ^{-1}\left\{-\tan \frac{\pi}{4}\right\} \\\\ &=\cos ^{-1}(-1) \\\\ &=\cos ^{-1}(\cos \pi) \\\\ &=\pi \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 20

Answer:
$\frac{1}{2}$
Given
$\cos \left(2 \sin ^{-1} \frac{1}{2}\right)$
Hint: Try to solve $\sin$ function.
Solution:
$\begin{aligned} \cos \left(2 \sin ^{-1} \frac{1}{2}\right) &=\cos \left(2 \times \frac{\pi}{6}\right)=\cos \left(\frac{\pi}{3}\right) \\ &=\frac{1}{2} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 21

Answer: $20^{\circ}$
Given: $\cos ^{-1}\left(\cos 350^{\circ}\right)-\sin ^{-1}\left(\sin 350^{\circ}\right)$
Hint: $\left[\sin \left(360^{\circ}-x\right)=-\sin x, \cos \left(360^{\circ}-x\right)=\cos x\right]$
Solution:
$\cos ^{-1}\left(\cos 350^{\circ}\right)-\sin ^{1}\left(\sin 350^{\circ}\right) \\$
$=\cos ^{-1}\left\{\cos \left(360^{\circ}-350^{\circ}\right)\right\}-\sin ^{-1}\left\{\sin \left(360^{\circ}-350^{\circ}\right)\right\} \\$
$=\cos ^{-1}\left\{\cos \left(10^{\circ}\right)\right\}-\sin ^{-1}\left\{\sin \left(-10^{\circ}\right)\right\} \\\\$
$=10^{\circ}-\left(-10^{\circ}\right) \\$
$= 20 ^\circ$

Inverse Trigonometric Functions exercise Very short answer question 22

Answer: $\frac{4}{5}$
Given: $\cos ^{2}\left(\frac{1}{2} \cos ^{-1} \frac{3}{5}\right) \\$
Hint: $\cos 2 x=2 \cos ^{2} x-1$
Solution:
$\text { Let, } y=\cos ^{-1}\left(\frac{3}{5}\right)\\$
$\cos y=\frac{3}{5} \\$
$\text { Now, } \cos ^{2}\left(\frac{1}{2} \cos ^{-1} \frac{3}{5}\right)=\cos ^{2}\left(\frac{1}{2} y\right) \\$
$=\frac{\cos y+1}{2} \quad\left[\therefore \cos 2 x=2 \cos ^{2} x-1\right] \\$
$=\frac{\frac{z}{5}+1}{2} \\$
$=\frac{\frac{8}{5}}{2} \\$
$=\frac{4}{5}$

Inverse Trigonometric Functions exercise Very short answer question 2

Answer: 1
Given:
$\tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{4}$
Hint: Try to separate the $\tan$ function into RHS, So that the variables gets free.
Solution:
$\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
Now,
$\begin{array}{l} \tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{4} \\\\ \tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\frac{\pi}{4} \\\\ \frac{x+y}{1-x y}=\tan \frac{\pi}{4} \\\\ \frac{x+y}{1-x y}=1 \\\\ x+y=1-x y \\\\ x+y+x y=1 \end{array}$

Inverse Trigonometric Functions exercise Very short answer question 24

Answer:$2 \pi-6$
Given: $\cos ^{-1}(\cos 6)$
Solution:
$\begin{aligned} \cos ^{-1}(\cos 6) &=\cos ^{-1}\{\cos (2 \pi-6)\} \\ &=2 \pi-6 \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 25

Answer: $\frac{7 \pi}{18}$
Given: $\sin ^{-1}\left(\cos \frac{\pi}{9}\right)$
Hint: $\cos x=\sin \left(\frac{\pi}{2}-x\right)$
Solution:
$\begin{aligned} \sin ^{-1}\left(\cos \frac{\pi}{9}\right) &=\sin ^{-1}\left\{\sin \left(\frac{\pi}{2}-\frac{\pi}{9}\right)\right\} \\ &=\sin ^{-1}\left\{\sin \left(\frac{7 \pi}{18}\right)\right\} \\ &=\frac{7 \pi}{18} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 26

Answer: 1
Given: $\sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right\}$
Hint: Try to solve the $\sin ^{-1}$ function.
Solution:
$\begin{aligned} \sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right\} &=\sin \left\{\frac{\pi}{3}-\left(\frac{-\pi}{6}\right)\right\} \\ &=\sin \left\{\frac{\pi}{3}+\frac{\pi}{6}\right\}=\sin \frac{\pi}{2} \\ &=1 \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 27

Answer:$-\frac{\pi}{4}$
Given: $\tan ^{-1}\left\{\tan \left(\frac{15 \pi}{4}\right)\right\}$
Hint: Try to solve the $\tan$ function first.
$\therefore \tan (4 \pi-x)=-\tan x$
Solution:
$\begin{aligned} \tan ^{-1}\left\{\tan \left(\frac{15 \pi}{4}\right)\right\} &=\tan ^{-1}\left\{\tan \left(4 \pi-\frac{\pi}{4}\right)\right\} \\ &=\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}\right)\right\} \\ &=\tan ^{1}\left\{\tan \left(-\frac{\pi}{4}\right)\right\} \\ &=\frac{-\pi}{4} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 28

Answer: $\pi$
Given: $2 \sin ^{-1} \frac{1}{2}+\cos ^{-1}\left(-\frac{1}{2}\right)$
Hint: Try to solve the $\sin ^{-1} \frac{1}{2} \ \ \& \ \ \cos ^{-1} \frac{1}{2}$ and add them.
Solution:
$\begin{aligned} 2 \sin ^{-1} \frac{1}{2}+\cos ^{-1}\left(-\frac{1}{2}\right) &=\sin ^{-1} 2 \times \frac{1}{2} \sqrt{1-\left(\frac{1}{2}\right)^{2}}+\cos ^{-1}\left(-\frac{1}{2}\right) \\ &=\sin ^{-1} \frac{\sqrt{3}}{2}+\cos ^{-1}\left(-\frac{1}{2}\right) \end{aligned}$
$\begin{array}{l} =\sin ^{-1}\left(\sin \frac{\pi}{3}\right)+\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right) \\\\ =\frac{\pi}{3}+\frac{2 \pi}{3} \\\\ =\pi \end{array}$

Inverse Trigonometric Functions exercise Very short answer question 29

Answer:$\frac{\pi}{4}$
Given:
Hint: $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$
Solution:
$\begin{aligned} \tan ^{-1} \frac{a}{b}-\tan ^{-1}\left(\frac{a-b}{a+b}\right) &=\tan ^{-1}\left(\frac{\frac{a}{b}-\frac{(a-b)}{a+b}}{1+\frac{a}{b}\left(\frac{a-b}{a+b}\right)}\right) \\\\ &=\tan ^{-1}\left(\frac{\frac{a^{2}+a b-a b+b^{2}}{b(a+b)}}{\frac{a b+b^{2}-a b+a^{2}}{b(a+b)}}\right) \\\\ &=\tan ^{-1}(1) \\\\ &=\tan ^{-1}\left(\tan \frac{\pi}{4}\right) \quad \therefore\left(\tan \frac{\pi}{4}=1\right) \\\\ &=\frac{\pi}{4} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 0

Answer: $\frac{3 \pi}{4}$
Given: $\cos ^{-1}\left(\cos \frac{5 \pi}{4}\right)$
Hint: Try to solve $\cos$ function first.
Solution:
$\cos ^{-1}\left(\cos \frac{5 \pi}{4}\right) \neq \frac{5 \pi}{4} \ \ as \ \ \frac{5 \pi}{4}$ doesn’t lie in between 0 and $5 \pi$
We, have
$\cos ^{-1}\left(\cos \frac{5 \pi}{4}\right)=\cos ^{-1}\left\{\cos \left(2 \pi-\frac{3 \pi}{4}\right)\right\}$
$\begin{array}{l} =\cos ^{-1}\left\{\cos \left(\frac{3 \pi}{4}\right)\right\} \\\\ =\frac{3 \pi}{4} \end{array}$

Inverse Trigonometric Functions exercise Very short answer question 1

Answer: $2 \sin ^{-1} x=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$
Given: To prove that
$\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x \\$
Hint:$x=\sin a$
Solution:
$\mathrm{LHS}=\sin ^{-1} 2 x \sqrt{1-x^{2}}$ Putting
$\begin {array} {ll} x=\sin a, we get \\\\ =\sin ^{-1}\left(2 \sin a \sqrt{1-\sin ^{2} a}\right) \\\\ =\sin ^{-1}(2 \sin a \cos a)\\\\ =\sin ^{-1}(\sin 2 a)\\\\ =2 a\\\\ =2 \sin ^{-1} x \quad[x=\sin a]\\\\ \end{array}$
$\therefore \sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x$

Inverse Trigonometric Functions exercise Very short answer question 2

Answer: $\frac{2 \pi}{5}$
Given:
$\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)$
Hint:
$\sin ^{-1}(\sin x)=x$
Solution:
$\begin{array}{l} \sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)=\sin ^{-1}\left\{\sin \left(\pi-\frac{3 \pi}{5}\right)\right\} \quad:\left[\left(\pi-\frac{3 \pi}{5}\right) \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\right] \\\\ \end{array}$
$\qquad \begin{array}{l} =\sin ^{-1}\left(\sin \frac{2 \pi}{5}\right) \\\\ =\frac{2 \pi}{5} \end{array}$

Inverse Trigonometric Functions exercise Very short answer question

Answer: $\sqrt 3$
Given:
$\tan ^{-1} \sqrt{3}+\cot ^{-1} x=\frac{\pi}{2}$
Hint:
$\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$
Solution:
$\begin{array}{l} \tan ^{-1}(\sqrt{3})+\cot ^{-1} x=\frac{\pi}{2} \\\\ \tan ^{-1}(\sqrt{3})=\frac{\pi}{2}-\cot ^{-1} x \\\\ \tan ^{-1}(\sqrt{3})=\tan ^{-1} x \\\\ \end{array}\\$
$x=\sqrt{3}$

Inverse Trigonometric Functions exercise Very short answer question 4

Answer:
$x=\frac{1}{3}$
Given: $\sin ^{-1}\left(\frac{1}{3}\right)+\cos ^{-1} x=\frac{\pi}{2}$
Hint:
$\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$
Solution:
\$\begin{array}{ll} \sin ^{-1}\left(\frac{1}{3}\right)+\cos ^{-1} x=\frac{\pi}{2} \\\\ \sin ^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{2}-\cos ^{-1} x \\\\ \sin ^{-1}\left(\frac{1}{3}\right)=\sin ^{-1} x \\\\ \end{array}$
$x=\frac{1}{2}$

Inverse Trigonometric Functions exercise Very short answer question 5

Answer:$\frac{-\pi}{2}$
Given: $\sin ^{-1}\left(\frac{1}{3}\right)-\cos ^{-1}\left(\frac{-1}{3}\right)$
Hint: $\cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2} \ \& \ \cos ^{-1}(-x)=\pi-\cos ^{-1} x$
Solution:
$\begin{aligned} \sin ^{-1}\left(\frac{1}{3}\right)-\cos ^{-1}\left(\frac{-1}{3}\right) &=\sin ^{-1}\left(\frac{1}{3}\right)-\left[\pi-\cos ^{-1}\left(\frac{1}{3}\right)\right] \\ &=\sin ^{-1}\left(\frac{1}{3}\right)+\cos ^{-1}\left(\frac{1}{3}\right)-\pi \\ &=\frac{\pi}{2}-\pi \\ &=-\frac{\pi}{2} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 6
Answer:$\frac{1}{2}$

Given:
$4 \sin ^{-1} x+\cos ^{-1} x=\pi$
Hint:
$\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$
Solution:
$\begin{array}{l} \therefore 4 \sin ^{-1} x+\cos ^{-1} x=\pi \\\\ 4 \sin ^{-1} x+\frac{\pi}{2}-\sin ^{-1} x=\pi \quad\left[\therefore \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right] \\\\ 3 \sin ^{-1} x=\frac{\pi}{2} \\\\ \sin ^{-1} x=\frac{\pi}{6} \\\\ x=\sin \frac{\pi}{6} \\\\ x=\frac{1}{2} \end{array}$

Inverse Trigonometric Functions exercise Very short answer question 7

Answer: $\frac{-\pi}{2}$
Given:
$x<0, y<0, x y=1$
$\tan ^{-1} x+\tan ^{-1} y=?$
Hint:
$\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
Solution:
$\begin{array}{l} x<0, y<0, \text { such that } x y=1 \\\\ \text { Let, } x=-a, y=-b \\ \end{array}$
$\qquad \therefore \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ \qquad \begin{aligned} \end{aligned}$
$\begin{aligned} &=\tan ^{-1}\left(\frac{-a-a}{1-1}\right) \\ &=\tan ^{-1}(-\infty) \\ &=\tan ^{-1}\left\{\tan \left(\frac{-\pi}{2}\right)\right\} \\ &=\frac{-\pi}{2} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 8

Answer:$\frac{-\pi}{3}$
Given:
$\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$
Hint: $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ is the range of the principal value branch of inverse sine function.
Solution:
$Let \ \ y=\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$
Then,
$\begin{array}{c} \sin y=-\frac{\sqrt{3}}{2}=\sin \left(-\frac{\pi}{3}\right) \\\\ \end{array}$
$\begin{array}{c} y=-\frac{\pi}{3} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\\\ \end{array}$
$\begin{array}{c} \therefore \sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3} \end{array}$

Inverse Trigonometric Functions exercise Very short answer question 9

Answer: $\frac{-\pi}{6}$
Given:
$\sin ^{-1}\left(-\frac{1}{2}\right)$
Hint:
$\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ is the range of the principal value branch of inverse sine function.
Solution:
$\begin{aligned} \text { Let } y=& \sin ^{-1}\left(-\frac{1}{2}\right) \\\\ & \sin y=-\frac{1}{2}=\sin \left(-\frac{\pi}{6}\right) \\\\ & y=-\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \end{aligned}$
Here,
$\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ is range of principal value of sin
$\sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}$

Inverse Trigonometric Functions exercise Very short answer question 40

Answer: $\pi$
Given:
$\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$
Hint:
$\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ is the range of the principal value branch of sine range of $\cos (0, \pi)$
Solution:
We have,
$\begin{aligned} \cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right) &=\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{\pi}{3}\right) \\ &=\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{\pi}{3}\right) \\ &=\frac{2 \pi}{3}+\frac{\pi}{3} \\ &= \pi \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 41

Answer: $\frac{5}{12}$
Given:
$\tan \left(2 \tan ^{-1} \frac{1}{5}\right)$
Hint:
Use formula
Solution:
$\begin{aligned} \tan \left(2 \tan ^{-1} \frac{1}{5}\right) &=\tan \left(\tan ^{-1} \frac{2 \times\left(\frac{1}{5}\right)}{1-\left(\frac{1}{5}\right)^{2}}\right) \\\\ &=\tan \left(\tan ^{-1} \frac{\frac{2}{5}}{\frac{24}{25}}\right) \\\\ &=\frac{5}{12} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 42

Answer: $\frac{11 \pi}{12}$
Given:
$\tan ^{-1} 1+\cos ^{-1}\left(\frac{-1}{2}\right)$
Hint:
Principal range of $\cos (0, \pi)$
Solution:
$\begin{aligned} \tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right) &=\tan ^{-1}\left(\tan \frac{\pi}{4}\right)+\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right) \\ &=\frac{\pi}{4}+\frac{2 \pi}{3} \\ &=\frac{11 \pi}{3} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 43

Answer: $\frac{\pi}{3}$
Given:$\tan ^{-1}\left[2 \sin \left(2 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right]$
Hint: Try to solve $\cos ^{-1}$ and then $\sin$ function.
Solution:
$\begin{aligned} \tan ^{-1}\left[2 \sin \left(2 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right] &=\tan ^{-1}\left\{2 \sin \left[\cos ^{-1} 2\left(\frac{\sqrt{3}}{2}\right)^{2}-1\right]\right.\\ &=\tan ^{-1}\left[2 \sin \left(\cos ^{-1} \frac{1}{2}\right)\right] = \frac{\pi}{3}\end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 44

Answer: $\frac{\pi}{2}$
Given
:$\tan ^{-1} \sqrt{3}+\cot ^{-1} \sqrt{3}$
Hint:
$\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$
Solution:
Using the hint we are able to solve this,
$\tan ^{-1} \sqrt{3}+\cot ^{-1} \sqrt{3}=\frac{\pi}{2}$

Inverse Trigonometric Functions exercise Very short answer question 45

Answer: $40 ^\circ$
Given: $\cos ^{-1}\left(\cos 680^{\circ}\right)$
Hint: Try to solve $\cos 680^{\circ}$ first, then separate and solve.
Solution:
$\begin{aligned} \cos ^{-1}\left(\cos 680^{\circ}\right) &=\cos ^{-1}\left[\cos \left(720^{\circ}-680^{\circ}\right)\right] \\ &=\cos ^{-1}\left(\cos 40^{\circ}\right) \\ &=40^{\circ} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 46

Answer: $\frac{2 \pi}{5}$
Given: $\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)$
Hint: Try to solve $\left(\sin \frac{3 \pi}{5}\right)$ first.
Solution:
$\begin{aligned} \sin ^{-1}\left(\sin \frac{3 \pi}{5}\right) &=\sin ^{-1}\left[\sin \left(\pi-\frac{2 \pi}{5}\right)\right] \\ &=\sin ^{-1}\left(\sin \frac{2 \pi}{5}\right) \\ &=\frac{2 \pi}{5} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 47

Answer: $\phi$
Given:
$\sec ^{-1}\left(\frac{1}{2}\right)$
Hint: The range of $\sec$ is (-1, 1) .
Solution:
The value of $\sec ^{-1}\left(\frac{1}{2}\right)$ is undefined as it is outside the range,
R-(-1, 1)

Inverse Trigonometric Functions exercise Very short answer question 48

Answer: $\frac{2 \pi}{3}$
Given: $\cos ^{-1}\left[\cos \left(4 \pi+\frac{2 \pi}{3}\right)\right]$
Hint: Try to solve $\cos \left(\frac{14 \pi}{3}\right)$ function.
Solution:
$\cos ^{-1}\left(\cos \frac{14 \pi}{3}\right)=\cos ^{-1}\left[\cos \left(4 \pi+\frac{2 \pi}{3}\right)\right]$
$\begin{array}{l} =\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right) \\\\ =\frac{2 \pi}{3} \end{array}$

Inverse Trigonometric Functions exercise Very short answer question 49

Answer: 0

Given: $\cos \left(\sin ^{-1} x+\cos ^{-1} x\right),|x| \leq 1$
Hint: $|x| \leq 1, x \in[-1,1]$
Solution:
$\begin {array}{ll} |x| \leq 1\\\\ \pm x \leq 1\\\\ x \leq 1 \ \ or \ \ x \geq-1\\\\ x \in[-1,1] \end {array}$
Now,
$\begin {array}{ll}\cos \left(\sin ^{-1} x+\cos ^{-1} x\right)=\cos \left(\frac{\pi}{2}\right) \\\\ =0\ \end {array}$

Inverse Trigonometric Functions exercise Very short answer question 50

Answer: 1
Given: $\tan \left(\frac{\sin ^{-1} x+\cos ^{-1} x}{2}\right)$ when $x=\frac{\sqrt{3}}{2}$
Hint: substitute x value into the function.
Solution:
$\begin{array} {ll} \tan \left(\frac{\sin ^{-1} x+\cos ^{-1} x}{2}\right)=\tan \left(\frac{\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)}{2}\right) \\\\ \therefore \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \\ \end{array}$
Hence,
$\tan \left(\frac{\sin ^{-1} x+\cos ^{-1} x}{2}\right)=\tan [(\pi / 4)=1$

Inverse Trigonometric Functions exercise Very short answer question 51

Answer: $\frac{\pi }{3}$
Given: $\sin ^{-1}\left[\cos \left(\sin ^{-1} \frac{1}{2}\right)\right]$
Hint: Try to solve $\sin ^{-1}$ function and its $\cos$ function.
Solution:
$\begin{aligned} \sin ^{-1}\left[\cos \left(\sin ^{-1} \frac{1}{2}\right)\right] &=\sin ^{-1}\left\{\cos \left[\sin ^{-1}\left(\sin \frac{\pi}{3}\right)\right]\right\} \\ &=\sin ^{-1}\left[\cos \left(\frac{\pi}{3}\right)\right] \\ &=\sin ^{-1}\left[\cos \left(\frac{\pi}{3}\right)\right] \\ &=\sin ^{-1}\left(\frac{1}{2}\right)=\sin ^{-1}\left(\sin \frac{\pi}{3}\right) \\ &=\frac{\pi}{3} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 52

Answer: $\emptyset$
Given:$\operatorname{cosec}^{-1}\left(\frac{\sqrt{3}}{2}\right)$
Hint: Range of principal value of $\operatorname{cosec} \ is \ (-1,1)$
Solution:
Range of principal value of is undefined as it’s outside the range
$(i.e., \(R-(-1,1)\rangle$

Inverse Trigonometric Functions exercise Very short answer question 53

Answer: $-\pi-\cot ^{-1} x\\$
Given
$\tan ^{-1}\left(\frac{1}{x}\right) \text { for } x<0$ in terms of $\cot ^{-1} x\$
Hint:
$\tan ^{-1}\left(\frac{1}{x}\right)=\tan ^{-1}\left(-\frac{1}{x}\right)$ for $x<0$
Solution:
$\begin{aligned} \tan ^{-1}\left(\frac{1}{x}\right) &=-\tan ^{-1}\left(\frac{1}{x}\right) \\ &=-\cot ^{-1} x \\ &=-\left(\pi-\cot ^{-1} x\right) \\ &=-\pi+\cot ^{-1} x \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 54

Answer: $\pi-\cot ^{-1} x$
Given:
$\cot ^{-1}(x) \ \ for \ \ all \ \ x \in R$ in terms of $\cot ^{-1}x$
Hint:
$\cot ^{-1}(-x)=\pi-\cot ^{-1} x$
Solution:
The value of $\cot ^{-1}(-x)=\pi-\cot ^{-1} x\) \ for \ all \ x \in \mathrm{R}$in terms of $\cot ^{-1} (x)$ is,
$=\pi-\cot ^{-1}(x)$

Inverse Trigonometric Functions exercise Very short answer question 55

Answer: $\frac{\sqrt{3}}{2}$
Given:
$\cos \left(\frac{\tan ^{-1} x+\cot ^{-1} x}{3}\right) when \ \ x=-\frac{1}{\sqrt{3}}$
Hint:
$\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$
Solution:
$\begin{aligned} \cos \left(\frac{\tan ^{-1} x+\cot ^{-1} x}{3}\right) &=\cos \left(\frac{\pi}{6}\right) \\ &=\frac{\sqrt{3}}{2} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 56

Answer: $\sqrt{3}$
Given:
$\cos \left(\tan ^{-1} x+\cot ^{-1} \sqrt{3}\right)=0$
Hint:
$\tan ^{-1} y+\cot ^{-1} y=\frac{\pi}{2}$
Solution:
$\begin{array}{l} \cos \left(\tan ^{-1} x+\cot ^{-1} \sqrt{3}\right)=\cos \left(\frac{\pi}{2}\right) \\\\ \tan ^{-1} x+\cot ^{-1} \sqrt{3}=\frac{\pi}{2} \\\\ x=\sqrt{3} \end{array}$

Inverse Trigonometric Functions exercise Very short answer question 57

Answer: $\frac{5 \pi}{6}$
Given:
$2 \sec ^{-1} 2+\sin ^{-1}\left(\frac{1}{2}\right)$
Hint: Try to convert or find the values of
$\sin ^{-1}\left(\frac{1}{2}\right), \sec ^{-1}(2)$
Solution:
$\begin{aligned} 2 \sec ^{-1} 2+\sin ^{-1}\left(\frac{1}{2}\right) &=2 \sec ^{1}\left(\sec \frac{\pi}{3}\right)+\sin ^{-1}\left(\sin \frac{\pi}{6}\right) \\ &=2 \times \frac{\pi}{3}+\frac{\pi}{6} \\ &=\frac{5 \pi}{6} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 58

Answer: $\frac{2}{5}$
Given:
$\cos \left(\sin ^{-1} \frac{2}{5}+\cos ^{-1} x\right)=0$
Hint: Try to convert the separate $\sin$ , $\cos$ function. So, that the variable get free.
Solution:
$\begin{array}{l} \cos \left(\sin ^{-1} \frac{2}{5}+\cos ^{-1} x\right)=0 \\\\ \cos \left(\sin ^{-1} \frac{2}{5}+\cos ^{-1} x\right)=\cos \left(\frac{\pi}{2}\right) \\\\ \sin ^{-1} \frac{2}{5}+\cos ^{-1} x=\frac{\pi}{2} \\\\ x=\frac{2}{5} c g f y \end{array}$

Inverse Trigonometric Functions exercise Very short answer question 59

Answer: $\frac{\pi}{6}$
Given:
$\cos ^{-1}\left(\cos ^{-1} \frac{3 \pi}{6}\right)$
Hint: To solve $\cos$ function we can get the solution.
Solution:
$\begin{aligned} \cos ^{-1}\left(\cos \frac{13 \pi}{6}\right) &=\cos ^{-1}\left[\cos \left(2 \pi+\frac{\pi}{6}\right)\right] \\ &=\cos ^{-1}\left(\cos \left(\frac{\pi}{6}\right)\right) \\ &=\frac{\pi}{6} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 60

Answer: $\frac{\pi}{8}$
Given
$\tan ^{-1}\left(\tan \frac{9 \pi}{8}\right)$
Hint: To solve $\tan$ function we can get the solution.
Solution:
$\begin{aligned} \tan ^{-1}\left(\tan \frac{9 \pi}{8}\right) &=\tan ^{-1}\left(\tan \left(\pi+\frac{\pi}{8}\right)\right) \\ &=\tan ^{-1}\left(\tan \left(\frac{\pi}{8}\right)\right) \\ &=\frac{\pi}{8} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 61

Answer: $\frac{-\pi}{2}$
Given: $\tan ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3})$
Hint: Range of
$\tan{ }^{-1} \ \ is \ \ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) ; \frac{\pi}{3} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
Range of
$\cot ^{-1} is (0, \pi) ; \frac{5 \pi}{6} \in[0, \pi]$
Solution:
$\begin{aligned} \tan ^{-1}(\sqrt{3})-\cot ^{-1}(-\sqrt{3}) &=\tan ^{-1}\left[\tan \left(\frac{\pi}{3}\right)\right]-\cot ^{-1}\left(\cot \frac{5 \pi}{6}\right) \\ &=\frac{\pi}{3}-\frac{5 \pi}{6} \\ &=-\frac{\pi}{2} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 62

Answer: $\frac{-\pi}{8}$
Given:
$\sin ^{-1}\left[\left(\sin -\frac{17 \pi}{8}\right)\right]$
Hint:
$\sin \left(\sin ^{-1} \theta\right)=\theta \ \ if \ \ -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$
Solution:
$\begin{aligned} \sin ^{-1}\left(\sin -\frac{17 \pi}{8}\right) \sin ^{-1}\left(-\sin \frac{17 \pi}{8}\right) &=\sin ^{-1}\left[-\sin \left(2 \pi+\frac{\pi}{8}\right)\right] \\ &=\sin ^{-1}\left(-\sin \frac{\pi}{8}\right) \\ &=\sin ^{-1}\left(\sin -\frac{\pi}{8}\right) \\ &=-\frac{\pi}{8} \end{aligned}$

Inverse Trigonometric Functions exercise Very short answer question 63

Answer: $\frac{3 \pi}{4}$
Given: $\cot ^{-1} 2\ \ and \ \ \cot ^{-1} 3$ , then third angle.
Hint: The sum of tree angles of a triangle is $\pi$
Solution:
$\begin{array}{l} \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{2}+\tan ^{-1} x=\pi \\\\ \tan ^{-1} \frac{\left(\frac{1}{3}+\frac{1}{2}\right)}{1-\frac{1}{6}}+\tan ^{-1} x=\pi \end{array}$

RD Sharma Class 12th VSAQ is based on Inverse Trigonometric Functions. To understand this chapter, students need to have a good idea about trigonometry. However, even if you don't have enough experience with trigonometry, you can still learn this chapter as the solution given by Career360 is for students to understand the chapter from scratch. RD Sharma Class 12th VSAQ can be considered small basic questions that the students can quickly go through.

In RD Sharma Class 12th Chapter 3 VSAQ, you will first evaluate the inverse functions and solve basic sums. This can be solved solely through fundamental operations. The higher-level questions which carry more marks are continued in the next section of this chapter.

There are a total of 62 VSAQs in this chapter which might seem quite overwhelming to students. Still, it is nothing to worry about as Class 12 RD Sharma Chapter 3 VSAQ Solution by Career360 has provided the best solutions which you can refer to and efficiently complete this section.VSAQs, although they carry fewer marks in exams but are essential where the weightage is concerned. It is through these VSAQs that you can learn crucial concepts for lengthy and complex problems. This is why you should always practice every one of them.

Practice, Planning, and Execution are the three factors that define an ideal Math student. As trigonometry consists of hundreds of sums, it is best to plan and divide those questions into parts. This method will be easier as you have to solve several sums per day while learning their concepts. Dividing up your work is very beneficial as it lessens the burden and gives the best output.

As Maths is such a vast subject getting a hold of all the concepts is pretty hard. Teachers can’t explain all the sums from the book, so you need to study these concepts beforehand to reduce the work needed afterward.

As far as trigonometry and inverse trigonometry are concerned, they contain hundreds of sums that are quite a lot for students to practice. This is where RD Sharma Class 12th Chapter 3 VSAQ shines. It includes all essential concepts with solutions for Inverse Trigonometry that can be accessible by everyone for free.

A majority of students have joined the team and started learning from Career360's material. You can also avail of this opportunity and download the material for free. This has excellent solutions which will help you score good marks in exams. So stay on top of your class and excel in trigonometric concepts with the help of RD Sharma Class 12 Chapter 3 VSAQ.

Chapter-wise RD Sharma Class 12 Solutions

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Frequently Asked Questions (FAQs)

1. Does this material have all chapters?

Yes, Career360 offers all the chapters on their website, available for free and accessible by students. You can learn from these solutions to score good marks in exams. For example, to get more insight on Inverse Trigonometry, you can download Class 12 RD Sharma Chapter 3 VSAQ Solution.

2. What are the applications of trigonometry?

These are some of the applications of trigonometry:

Oceanography - To calculate tide heights.
Calculus
Physics - Sine and Cos functions are used as wave representations.
Height calculation - To Measure the height of buildings or mountains.

3. Who invented Inverse Trigonometry?

Daniel Bernoulli introduced inverse trigonometry. This concept was used to determine the angle if two sides of a triangle are given. For more information on Inverse Trigonometry, check out RD Sharma Class 12 Solutions Inverse Trigonometric Functions VSAQ

4. Are NCERT and RD Sharma books similar?

NCERT books for maths are good for general preparation, but RD Sharma books are relatively better to dive deep into the concepts. They are widely considered by teachers and students all over the country and are excellent means of studying for exams. To find the RD Sharma material for Inverse Trigonometry, you can study from RD Sharma Class 12 Solutions Chapter 3 VSAQ.

5. Where is inverse trigonometry used in real life?

nverse trigonometry is a concept through which we can determine the third side of a triangle if two are known. This has its applications across engineering physics, geometry, and navigation on land and sea. To get more information about Inverse Trigonometry, you can refer to RD Sharma Class 12 Solutions Inverse Trigonometric Functions VSAQ.