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RD Sharma Class 12 Solutions Chapter 3 Inverse Trigonometric Functions - Other Exercise
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
$a< x< b$ .
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 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 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 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 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,
$\begin{array}{c} \sin y=-\frac{\sqrt{3}}{2}=\sin \left(-\frac{\pi}{3}\right) \\\\ \end{array}$ $\begin{array}{c} y=-\frac{\pi}{3} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\\\ \end{array}$ $\begin{array}{c} \therefore \sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3} \end{array}$ 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 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 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 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 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.
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