Exercise 4.6 - Chapter 4 Inverse Trigonometric Functions 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $4.6$
Choose the correct or the most suitable answer from the given four alternatives.
Question 1.
The value of $\sin ^{-1}(\cos x), 0 \leq x \leq \pi$ is
(a) $\pi-x$
(b) $\mathrm{x}-\frac{\pi}{2}$
(c) $\frac{\pi}{2}-x$
(d) $\pi-x$
Solution:
(c) $\frac{\pi}{2}-\mathrm{x}$
Hint:
We know $\cos x=\sin \left(\frac{\pi}{2}-x\right)$
So $\sin ^{-1}(\cos x)=\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-x\right)\right]=\frac{\pi}{2}-x$
Question 2.
If $\sin ^{-1}+\sin ^{-1} y=\frac{2 \pi}{3}$; then $\cos ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{y}$ is equal to
(a) $\frac{2 \pi}{3}$
(b) $\frac{\pi}{3}$
(c) $\frac{\pi}{6}$
(d) $\pi$
Solution:
(b) $\frac{\pi}{3}$
Hint:
We know $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$
and $\sin ^{-1} y+\cos ^{-1} y=\frac{\pi}{2}$
Adding them
so $\left(\sin ^{-1} x+\sin ^{-1} y\right)+\left(\cos ^{-1} x+\cos ^{-1} y\right)=\pi$
i.e. $\quad \frac{2 \pi}{3}+\left(\cos ^{-1} x+\cos ^{-1} y\right)=\pi$
$\Rightarrow \quad \cos ^{-1} x+\cos ^{-1} y=\pi-\frac{2 \pi}{3}=\frac{\pi}{3}$

Question 3.
$
\sin ^{-1} \frac{3}{5}-\cos ^{-1} \frac{12}{13}+\sec ^{-1} \frac{5}{3}-\operatorname{cosec}^{-1} \frac{13}{12}=
$
(a) $2 \pi$
(b) $\pi$
(c) 0
(d) $\tan ^{-1} \frac{12}{65}$
Solution:
(c) 0
Hint:
We know $\sin ^{-1} \frac{3}{5}+\cos ^{-1} \frac{3}{5}=\frac{\pi}{2}$
and $\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{12}{13}=\frac{\pi}{2}$
so
$
\frac{\pi}{2}-\frac{\pi}{2}=0
$
Question 4.
If $\sin ^{-1} x=2 \sin ^{-1} \alpha$ has a solution, then
(a) $|\alpha| \leq \frac{1}{\sqrt{2}}$
(b) $|\alpha| \geq \frac{1}{\sqrt{2}}$
(c) $|\alpha|<\frac{1}{\sqrt{2}}$
(d) $|\alpha|>\frac{1}{\sqrt{2}}$
Solution:
(a) $|\alpha| \leq \frac{1}{\sqrt{2}}$
Hint:

$\begin{aligned}
& \sin ^{-1} x=2 \sin ^{-1} \alpha \\
& \text { Range of } \sin ^{-1} x \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\
& \text { i.e. } \quad-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2} \\
& \Rightarrow \quad-\frac{\pi}{2} \leq 2 \sin ^{-1} \alpha \leq \frac{\pi}{2} \quad \Rightarrow-\frac{\pi}{4} \leq \sin ^{-1} \alpha \leq \frac{\pi}{4} \\
& \Rightarrow \quad \sin ^{-1}\left(-\frac{\pi}{4}\right) \leq \alpha \leq \sin \left(\frac{\pi}{4}\right) \quad \Rightarrow-\frac{1}{\sqrt{2}} \leq \alpha \leq \frac{1}{\sqrt{2}} \\
& \Rightarrow \quad|\alpha|=\frac{1}{\sqrt{2}} \\
&
\end{aligned}$

Question 5.
$\sin ^{-1}(\cos x)=\frac{\pi}{2}-x$ is valid for
(a) $-\pi \leq x \leq 0$
(b) $0 \leq \mathrm{x} \leq \pi$
(c) $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$
(d) $-\frac{\pi}{4} \leq \mathrm{x} \leq \frac{3 \pi}{4}$
Solution:
(b) $0 \leq \mathrm{x} \leq \pi$
Question 6.
If $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2}$, the value of $x^{2017}+y^{2018}+z^{2019}-\frac{9}{x^{101}+y^{101}+z^{101}}$ is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(a) 0
Hint:
The maximum value of $\sin ^{-1} x$ is $\frac{\pi}{2}$ and $\sin ^{-1} 1=\frac{\pi}{2}$
Here it is given that
$
\begin{aligned}
& \sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2} \\
& \Rightarrow x=y=z=1 \\
& \text { and so } 1+1+1-\frac{9}{1+1+1}=3-3=0
\end{aligned}
$
Question 7.
If $\cot ^{-1} x=\frac{2 \pi}{5}$ for some $x \in R$, the value of $\tan ^{-1} x$ is $\ldots \ldots \ldots .$.
(a) $-\frac{\pi}{10}$
(b) $\frac{\pi}{5}$
(c) $\frac{\pi}{10}$
(d) $-\frac{\pi}{5}$
Solution:
(c) $\frac{\pi}{10}$
Hint:
Given $\cot ^{-1} x=\frac{2 \pi}{5} \quad \Rightarrow x=\cot \frac{2 \pi}{5}$
We know $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$
SO
$
\tan ^{-1} x=\frac{\pi}{2}-\frac{2 \pi}{5}=\frac{\pi}{10}
$

Question 8.
The domain of the function defined by $\mathrm{f}(\mathrm{x})=\sin ^{-1} \sqrt{x-1}$ is
(a) $[1,2]$
(b) $[-1,1]$
(c) $[0,1]$
(d) $[-1,0]$
Solution:
(a) $[1,2]$
Hint:
The domain for $\sin ^{-1} \mathrm{x}$ is $[0,1]$
So $\sqrt{x-1}=0 \Rightarrow \mathrm{x}-1=0 \Rightarrow \mathrm{x}=1$
$\sqrt{x-1}=1 \Rightarrow \mathrm{x}-1=0 \Rightarrow \mathrm{x}=2$
$\therefore$ The domain is $[1,2]$
Question 9.
If $x=\frac{1}{5}$, the value of $\cos \left(\cos ^{-1} x+2 \sin ^{-1} x\right)$ is
(a) $-\sqrt{\frac{24}{25}}$
(b) $\sqrt{\frac{24}{25}}$
(c) $\frac{1}{5}$
(d) $-\frac{1}{5}$
Solution:
(d) $-\frac{1}{5}$
Hint:
$
\text { Now } \begin{aligned}
\cos \left[\cos ^{-1} \frac{1}{5}+2 \sin ^{-1} \frac{1}{5}\right] & =\cos \left[\cos ^{-1} \frac{1}{5}+\sin ^{-1} \frac{1}{5}+\sin ^{-1} \frac{1}{5}\right] \\
& =\cos \left[\frac{\pi}{2}+\sin ^{-1} \frac{1}{5}\right]=\cos \left[\frac{\pi}{2}+\frac{\pi}{2}-\cos ^{-1} \frac{1}{5}\right] \\
& =\cos \left[\pi-\cos ^{-1} \frac{1}{5}\right]=\cos \cos ^{-1}\left(-\frac{1}{5}\right)=-\frac{1}{5}
\end{aligned}
$

Question 10.
$\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)$ is equal to
(a) $\frac{1}{2} \cos ^{-1}\left(\frac{3}{5}\right)$
(b) $\frac{1}{2} \sin ^{-1}\left(\frac{3}{5}\right)$
(c) $\frac{1}{2} \tan ^{-1}\left(\frac{3}{5}\right)$
(d) $\tan ^{-1}\left(\frac{1}{2}\right)$
Solution:
(d) $\tan ^{-1} \frac{1}{2}$
Hint:
$
\begin{aligned}
\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9} & =\tan ^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4} \times \frac{2}{9}}\right)=\tan ^{-1}\left(\frac{\frac{9+8}{36}}{1-\frac{2}{36}}\right) . \\
& =\tan ^{-1}\left(\frac{17}{34}\right)=\tan ^{-1}\left(\frac{1}{2}\right)
\end{aligned}
$
Question 11.
If the function $f(x)=\sin ^{-1}\left(x^2-3\right)$, then $x$ belongs to
(a) $[1,-1]$
(b) $[\sqrt{2}, 2]$
(c) $[-2,-\sqrt{2}] \cup[\sqrt{2}, 2]$
(d) $[-2,-\sqrt{2}] \cap[\sqrt{2}, 2]$
Solution:
(c) $[-2,-\sqrt{2}] \cup[\sqrt{2}, 2]$
Hint:
$
f(x)=\sin ^{-1}\left(x^2-3\right)
$
Domain of $\sin ^{-1}(\mathrm{x})$ is $[-1,1]$
$
\begin{aligned}
& \Rightarrow-1 \leq x^2-3 \leq 1 \Rightarrow 2 \leq x^2 \leq 4 \\
& \Rightarrow \sqrt{2} \leq x \leq 2 \Rightarrow \sqrt{2} \leq|x| \leq 2 \\
& x \in[-2,-\sqrt{2}] \cup[\sqrt{2}, 2]
\end{aligned}
$
Question 12.
If $\cot ^{-1} 2$ and $\cot ^{-1} 3$ are two angles of a triangle, then the third angle is
(a) $\frac{\pi}{4}$
(b) $\frac{3 \pi}{4}$
(c) $\frac{\pi}{6}$
(d) $\frac{\pi}{3}$
Solution:
(b) $\frac{3 \pi}{4}$

Hint:
Let $\cot ^{-1} 2=\theta_1 \quad \Rightarrow \cot \theta_1=2$ so $\quad \tan \theta_1=\frac{1}{2}$
Let $\cot ^{-1} 3=\theta_2 \quad \Rightarrow \cot \theta_2=3$ so $\quad \tan \theta_2=\frac{1}{3}$
$
\begin{aligned}
& \tan \left(\theta_1+\theta_2\right)=\frac{\tan \theta_1+\tan \theta_2}{1-\tan \theta_1 \tan \theta_2}=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}} \\
& \Rightarrow \quad \tan \left(\theta_1+\theta_2\right)=\frac{\frac{3+2}{6}}{1-\frac{1}{6}}=\frac{\frac{5}{6}}{\frac{6}{6}}=1 \quad \Rightarrow \theta_1+\theta_2=\frac{\pi}{4}
\end{aligned}
$
Let $\theta_3$ be the third angle
$\mathrm{SO}$
$
\begin{aligned}
& \theta_1+\theta_2+\theta_3=180=\pi \\
& \Rightarrow \quad \theta_3=\pi-\frac{\pi}{4}=\frac{3 \pi}{4} \\
&
\end{aligned}
$

Question 13.
$\sin ^{-1}\left(\tan \frac{\pi}{4}\right)-\sin ^{-1}\left(\sqrt{\frac{3}{x}}\right)=\frac{\pi}{6}$. Then $\mathrm{x}$ is a root of the equation
(a) $x^2-x-6=0$
(b) $x^2-x-12=0$
(c) $x^2+x-12=0$
(d) $x^2+x-6=0$
Solution:
(b) $x^2-x-12=0$
Hint:
$
\begin{aligned}
& \sin ^{-1}(1)-\sin ^{-1}\left(\frac{\sqrt{3}}{\sqrt{x}}\right)=\frac{\pi}{6} \\
& \Rightarrow \quad \frac{\pi}{2}-\sin ^{-1} \frac{\sqrt{3}}{\sqrt{x}}=\frac{\pi}{6} \quad \Rightarrow \sin ^{-1} \frac{\sqrt{3}}{\sqrt{x}}=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3} \\
& \Rightarrow \quad \frac{\sqrt{3}}{\sqrt{x}}=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2} \quad \Rightarrow \sqrt{x}=\frac{2 \sqrt{3}}{\sqrt{3}}=2 \quad \Rightarrow x=4 \\
&
\end{aligned}
$
Question 14.
$
\sin ^{-1}\left(2 \cos ^2 x-1\right)+\cos ^{-1}\left(1-2 \sin ^2 x\right)=
$

(a) $\frac{\pi}{2}$.
(b) $\frac{\pi}{3}$
(c) $\frac{\pi}{4}$
(d) $\frac{\pi}{6}$
Solution:
(a) $\frac{\pi}{2}$
Hint:
$
\begin{aligned}
& 2 \cos ^2 x-1=\cos 2 x \\
& 1-2 \sin ^2 x=\cos 2 x \\
& \therefore \sin ^{-1} x(\cos 2 x)+\cos ^{-1}(\cos 2 x)=\frac{\pi}{2}\left(\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right)
\end{aligned}
$
Question 15.
If $\cot ^{-1}(\sqrt{\sin \alpha})+\tan ^{-1}(\sqrt{\sin \alpha})=\mathrm{u}$, then $\cos 2 \mathrm{u}$ is equal to
(a) $\tan ^2 \alpha$
(b) 0
(c) $-1$
(d) $\tan 2 \alpha$
Solution:
(c) $-1$
Hint:
$
\begin{aligned}
& \cot ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{x}=\frac{\pi}{2} \Rightarrow \mathrm{u}=\frac{\pi}{2} \text { so } 2 \mathrm{u}=\pi \\
& \therefore \cos 2 \mathrm{u}=\cos \pi=-1
\end{aligned}
$
Question 16.
If $|x| \leq 1$, then $2 \tan ^{-1} x-\sin ^{-1} \frac{2 x}{1+x^2}$ is equal to
(a) $\tan ^{-1} x$
(b) $\sin ^{-1} x$
(c) 0
(d) $\pi$
Solution:
(c) 0
Hint:
Let $\mathrm{x}=\tan \theta$ so $\frac{2 x}{1+x^2}=\sin 2 \theta$.
Now $2 \tan ^{-1}(\tan \theta)-\sin ^{-1}(\sin 2 \theta)=2 \theta-2 \theta=0$
Question 17.
The equation $\tan ^{-1} x-\cot ^{-1} x=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$ has
(a) no solution
(b) unique solution
(c) two solutions
(d) infinite number of solutions
Solution:
(b) unique solution

Hint:
$
\begin{aligned}
\tan ^{-1} x-\cot ^{-1} x & =\tan ^{-1} \frac{1}{\sqrt{3}} \\
\Rightarrow \quad \tan ^{-1} x-\left(\frac{\pi}{2}-\tan ^{-1} x\right) & =\frac{\pi}{6} \\
\tan ^{-1} x+\tan ^{-1} x & =\frac{\pi}{2}+\frac{\pi}{6} \\
2 \tan ^{-1} x & =\frac{2 \pi}{3} \quad \Rightarrow x=\tan \frac{\pi}{3}=\sqrt{3}
\end{aligned}
$
$\Rightarrow$ It has a unique Solution

Question 18.
If $\sin ^{-1} x+\cot ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2}$, then $x$ is equal to
(a) $\frac{1}{2}$
(b) $\frac{1}{\sqrt{5}}$
(c) $\frac{2}{\sqrt{5}}$
(d) $\frac{\sqrt{3}}{2}$
Solution:
(b) $\frac{1}{\sqrt{5}}$
Hint:
$
\begin{aligned}
& \sin ^{-1} x+\cot ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2} \\
& \Rightarrow \quad \sin ^{-1} x=\frac{\pi}{2}-\cot ^{-1} \frac{1}{2}=\tan ^{-1} \frac{1}{2} \\
& \sin ^{-1} x=\sin ^{-1}\left(\frac{\frac{1}{2}}{\sqrt{1+\left(\frac{1}{2}\right)^2}}\right) \\
& \Rightarrow \quad \sin ^{-1} x=\sin ^{-1}\left(\frac{\frac{1}{2}}{\frac{\sqrt{5}}{2}}\right)=\frac{1}{\sqrt{5}} \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right) \\
& \Rightarrow \quad x=\frac{1}{\sqrt{5}} \\
&
\end{aligned}
$
Question 19.
If $\sin ^{-1} \frac{x}{5}+\operatorname{cosec}^{-1} \frac{5}{4}=\frac{\pi}{2}$, then the value of $\mathrm{x}$ is ........
(a) 4

(b) 5
(c) 2
(d) 3
Solution:
(d) 3
Hint:

$
\begin{aligned}
& \sin ^{-1} \frac{x}{5}+\operatorname{cosec}^{-1} \frac{5}{4}=\frac{\pi}{2} \\
& \Rightarrow \quad \sin ^{-1} \frac{x}{5}=\frac{\pi}{2}-\operatorname{cosec}^{-1} \frac{5}{4}=\sec ^{-1} \frac{5}{4} \\
& \Rightarrow \quad \sin ^{-1} \frac{x}{5}=\sec ^{-1} \frac{5}{4}=\cos ^{-1} \frac{4}{5} \\
& \text { Let } \quad \sin ^{-1} \frac{x}{5}=\theta \\
& \Rightarrow \quad \sin \theta=\frac{x}{5} \\
& \text { so } \quad \cos \theta=\frac{\sqrt{25-x^2}}{5} \\
& \Rightarrow \quad \theta=\cos ^{-1} \frac{\sqrt{25-x^2}}{5}=\cos ^{-1} \frac{4}{5} \text { (given) } \\
& \Rightarrow \quad \frac{\sqrt{25-x^2}}{5}=\frac{4}{5} \quad \Rightarrow \sqrt{25-x^2}=4 \\
& \Rightarrow \quad 25-x^2=16 \quad \Rightarrow x^2=9 \quad \Rightarrow x=3 \\
&
\end{aligned}
$

Question 20.
$\sin \left(\tan ^{-1}\right),|x|<1$ is equal to
(a) $\frac{x}{\sqrt{1-x^2}}$
(b) $\frac{1}{\sqrt{1-x^2}}$
(c) $\frac{1}{\sqrt{1+x^2}}$
(d) $\frac{x}{\sqrt{1+x^2}}$
Solution:
(d) $\frac{x}{\sqrt{1+x^2}}$
Hint:
$
\begin{aligned}
\sin \left(\tan ^{-1} x\right) & =\sin \left(\sin ^{-1} \frac{x}{\sqrt{1+x^2}}\right)=\frac{x}{\sqrt{1+x^2}} \\
& =\frac{x}{\sqrt{1+x^2}}
\end{aligned}
$