Ch.2 : Inverse Trigonometric Functions MCQ's Solutions - Inverse Trigonometry - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ch.2 : Inverse Trigonometric Functions MCQ's & Case Study Questions - PDF Download Here and Check Solutions Below 

1) Answer:  (a)
Explanation:
Given that,
$$
\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}
$$
Therefore, $\left(\frac{\pi}{2}-\cos ^{-1} x\right)+\left(\frac{\pi}{2}-\cos ^{-1} y\right)=\frac{\pi}{2}$
$$
\cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2}
$$

2) Answer: (b)
Explanation:
$$
\begin{aligned}
&\sin ^{-1} x-\cos ^{-1} x=\sin ^{-1}\left(\frac{1}{2}\right) \\
&\Rightarrow \sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6} \\
&\Rightarrow \sin ^{-1} x-\left(\frac{\pi}{2}-\sin ^{-1} x\right)=\frac{\pi}{6} \\
&\Rightarrow 2 \sin ^{-1} x=\frac{\pi}{6}+\frac{\pi}{2}=\frac{2 \pi}{3} \\
&\Rightarrow \sin ^{-1} x=\frac{\pi}{3} \\
&\Rightarrow x=\frac{\sqrt{3}}{2}
\end{aligned}
$$
Hence, there is only one solution.

3) Answer: (a)
Explanation:
We know that,
$$
\cot ^{-1}\left(\frac{x y+1}{x-y}\right)=\cot ^{-1} x-\cot ^{-1} y
$$
Now,
$$
\begin{aligned}
&=\cot ^{-1}\left(\frac{a b+1}{a-b}\right)+\cot ^{-1}\left(\frac{b c+1}{b-c}\right)+\cot ^{-1}\left(\frac{c a+1}{c-a}\right) \\
&=\cot ^{-1} a-\cot ^{-1} b+\cot ^{-1} b-\cot ^{-1} c+\cot ^{-1} c-\cot ^{-1} a \\
&=0
\end{aligned}
$$

4) Answer: (d)
Explanation:
$$
\begin{aligned}
&\sin ^{-1} \sin \left(-600 \times \frac{\pi}{180}\right) \\
&=\sin ^{-1} \sin \left(\frac{-10 \pi}{3}\right) \\
&=\sin ^{-1}\left[\sin \left(4 \pi-\frac{2 \pi}{3}\right)\right] \\
&=\sin ^{-1}\left[\sin \left(\frac{2 \pi}{3}\right)\right] \\
&=\sin ^{-1}\left[\sin \left(\pi-\frac{\pi}{3}\right)\right] \\
&=\sin ^{-1}\left[\sin \left(\frac{\pi}{3}\right)\right] \\
&=\frac{\pi}{3}
\end{aligned}
$$

5) Answer: (b)
Explanation:
The Principal value branch of $\operatorname{Sec}^{-1} \mathrm{x}$ is $[0, \pi]-\left\{\frac{\pi}{2}\right\}$

6) Answer: (a)
Explanation:
$$
\begin{aligned}
&=4 \tan ^{-1}\left(\frac{1}{5}\right)-\tan ^{-1}\left(\frac{1}{239}\right) \\
&=2\left(2 \tan ^{-1}\left(\frac{1}{5}\right)\right)-\tan ^{-1}\left(\frac{1}{239}\right) \\
&=2\left(2 \tan ^{-1}\left(\frac{2 \cdot \frac{1}{5}}{1-\frac{1}{25}}\right)\right)-\tan ^{-1}\left(\frac{1}{239}\right) \\
&=2\left(2 \tan ^{-1}\left(\frac{5}{12}\right)\right)-\tan ^{-1}\left(\frac{1}{239}\right) \\
&=\tan ^{-1}\left(\frac{120}{119}\right)-\tan ^{-1}\left(\frac{1}{239}\right) \\
&=\tan ^{-1}\left(\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119} \cdot \frac{1}{239}}\right) \\
&=\tan ^{-1}(1)=\frac{\pi}{4} .
\end{aligned}
$$

7) Answer : (c)
Explanation:
$$
\begin{aligned}
&\tan \left(\cos ^{-1} \frac{3}{5}+\tan ^{-1} \frac{1}{4}\right) \\
&=\tan \left(\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{4}\right) \\
&=\tan \left(\tan ^{-1}\left(\frac{\frac{4}{3}+\frac{1}{4}}{1-\frac{4}{3} \cdot \frac{1}{4}}\right)\right) \\
&=\tan \left(\tan ^{-1}\left(\frac{19}{8}\right)\right) \\
&=\frac{19}{8} .
\end{aligned}
$$

8) Answer: $\quad$ (a)
Explanation:
We know that $\tan ^{-1}\left(\frac{x-y}{1+x y}\right)=\tan ^{-1} \mathrm{x}-\tan ^{-1} \mathrm{y}$
$$
\begin{aligned}
&=\tan ^{-1}\left(\frac{a b+1}{a-b}\right)+\tan ^{-1}\left(\frac{b c+1}{b-c}\right)+\tan ^{-1}\left(\frac{c a+1}{c-a}\right) \\
&=\tan ^{-1} \mathrm{a}-\tan ^{-1} \mathrm{~b}+\tan ^{-1} \mathrm{~b}-\tan ^{-1} \mathrm{c}+\tan ^{-1} \mathrm{c}-\tan ^{-1} \mathrm{a} \\
&=0
\end{aligned}
$$

9) Answer: (a)
Explanation:
$$
\begin{aligned}
&\sin \left[\cot ^{-1}\left(\cos \frac{\pi}{4}\right)\right] \\
&=\sin \left[\cot ^{-1} \frac{1}{\sqrt{2}}\right] \\
&=\sin \left[\sin ^{-1} \sqrt{\frac{2}{3}}\right] \\
&=\sqrt{\frac{2}{3}}
\end{aligned}
$$

10) Answer: (b)
Explanation:
$$
\begin{aligned}
&=\tan ^{-1} x+\tan ^{-1}\left(\frac{1}{x}\right) \\
&=\tan ^{-1} x+\cot ^{-1} x \\
&=\frac{\pi}{2}
\end{aligned}
$$

11) Answer: (a)
Explanation:
$$
\begin{aligned}
\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right) &=\sin ^{-1}\left(-\sin \frac{\pi}{3}\right) \\
&=-\sin ^{-1}\left(\sin \frac{\pi}{3}\right) \\
&=-\frac{\pi}{3}
\end{aligned}
$$ 

12) Answer: (c)
Explanation:
Since, the domain of $\sin ^{-1} x$ is $[-1,1]$
and domain of $\sec ^{-1} \mathrm{x}$ is $\mathrm{R}-(-1,1), \mathrm{D}_{\mathrm{f}}=\{-1,1\}$

13) Answer: (a)
Explanation:
$$
\begin{aligned}
&\cos ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{y}=2 \pi \\
&\Rightarrow \frac{\pi}{2}-\sin ^{-1}(\mathrm{x})+\frac{\pi}{2}-\sin ^{-1}(\mathrm{y})=2 \pi,\left[\because \sin ^{-1}(\mathrm{x})+\cos ^{-1}(\mathrm{x})=\frac{\pi}{2}\right] \\
&\Rightarrow-\sin ^{-1}(\mathrm{x})-\sin ^{-1}(\mathrm{y})=\pi \\
&\Rightarrow \sin ^{-1}(\mathrm{x})+\sin ^{-1}(\mathrm{y})=-\pi
\end{aligned}
$$

14) Answer: (a)
Explanation:
$$
3 \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)-4 \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+2 \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}
$$
Put $x=\tan \theta$
$$
\begin{aligned}
&3 \sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)-4 \cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)+2 \tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right)=\frac{\pi}{3} \\
&3 \sin ^{-1}(\sin 2 \theta)-4 \cos ^{-1}(\cos 2 \theta)+2 \tan ^{-1}(\tan 2 \theta)=\frac{\pi}{3} \\
&3(2 \theta)-4(2 \theta)+2(2 \theta)=\frac{\pi}{3} \\
&\Rightarrow 2 \theta=\frac{\pi}{3} \\
&\Rightarrow \theta=\frac{\pi}{6} \\
&\Rightarrow \tan ^{-1} \mathrm{x}=\frac{\pi}{6} \\
&\Rightarrow \mathrm{x}=\tan \left(\frac{\pi}{6}\right)=\frac{1}{\sqrt{3}}
\end{aligned}
$$

15) Answer: $\quad$ (a)
Explanation:
We know that principal value branch of $\cos ^{-1}$ is $[0, \pi]$
and $\frac{3 \pi}{4} \notin[0, \pi] .$ But $\left(2 \pi-\frac{5 \pi}{4}\right) \in[0, \pi]$
$$
\begin{aligned}
\cos ^{-1}\left(\cos \frac{5 \pi}{4}\right) &=\cos ^{-1}\left(\cos \left(\left(2 \pi-\frac{5 \pi}{4}\right)\right)\right) \\
=& \cos ^{-1}\left(\cos \frac{3 \pi}{4}\right)=\frac{3 \pi}{4}
\end{aligned}
$$

16) Answer: (c)
Explanation:
$$
\begin{aligned}
&\sin ^{-1} x+\cos ^{-1} y=\frac{2 \pi}{3} \\
&\Rightarrow \frac{\pi}{2}-\cos ^{-1} x+\frac{\pi}{2}-\cos ^{-1} y=\frac{2 \pi}{3} \\
&\Rightarrow \pi-\left(\cos ^{-1} x+\cos ^{-1} y\right)=\frac{2 \pi}{3} \\
&\cos ^{-1} x+\cos ^{-1} y=\pi-\frac{2 \pi}{3}=\frac{\pi}{3}
\end{aligned}
$$

17) Answer: (b)
Explanation:
By Converting the ratios to other ratios, We get,
$$
\begin{aligned}
&\cot ^{-1}\left(\frac{5}{3}\right)=\tan ^{-1}\left(\frac{3}{5}\right) \text { and } \cos ^{-1}\left(\frac{4}{5}\right)=\tan ^{-1}\left(\frac{3}{4}\right) \\
&=\cot ^{-1}\left(\frac{5}{3}\right)+\cos ^{-1}\left(\frac{4}{5}\right) \\
&=\tan ^{-1}\left(\frac{3}{5}\right)+\tan ^{-1}\left(\frac{3}{4}\right) \\
&=\left(\frac{\frac{3}{5}+\frac{3}{4}}{1-\left(\frac{3}{5}\right)\left(\frac{3}{4}\right)}\right) \\
&=\tan ^{-1}\left(\frac{27}{11}\right)
\end{aligned}
$$

18) Answer: (d)
Explanation:
$$
\begin{aligned}
&\text { Put, } \tan ^{-1} \mathrm{x}=\theta \\
&\Rightarrow \mathrm{x}=\tan \theta \\
&\Rightarrow \tan \theta=\frac{x}{1}=\frac{p e r p}{\text { base }} \\
&\Rightarrow \cos \theta=\frac{\text { base }}{\text { hyp. }}=\frac{1}{\sqrt{1+x^{2}}}
\end{aligned}
$$

19) Answer : $\mid(b)$
Explanation:
$$
\begin{aligned}
&=\sin ^{-1}\left(\frac{3}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right) \\
&=\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{1}{7}\right) \\
&=\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{1}{7}}{1-\left(\frac{3}{4}\right)\left(\frac{1}{7}\right)}\right)=\tan ^{-1}(1)=\frac{\pi}{4}
\end{aligned}
$$

20) Answer: (b)
Explanation:
Let $\sin ^{-1} 2 \mathrm{x}=\theta$
So that $2 \mathrm{x}=\sin \theta$
Now $-1 \leq \sin \theta \leq 1$, i.e., $-1 \leq 2 \mathrm{x} \leq 1$ which gives $-\frac{1}{2} \leq \mathrm{x} \leq \frac{1}{2}$

21) Answer: (c)
Explanation:
$$
\begin{aligned}
&\cos ^{-1} \mathrm{x}+\sin ^{-1}\left(\frac{x}{2}\right)=\frac{\pi}{6} \\
&\Rightarrow \sin ^{-1}\left(\frac{x}{2}\right)=\frac{\pi}{6}-\cos ^{-1} \mathrm{x} \\
&\Rightarrow \frac{x}{2}=\sin \left(\left(\frac{\pi}{6}\right)-\cos ^{-1} x\right) \\
&\Rightarrow \frac{x}{2}=\frac{1}{2} \mathrm{x}-\frac{\sqrt{3}}{2} \sin \left(\cos ^{-1} \mathrm{x}\right) \\
&\Rightarrow \sin \left(\cos ^{-1} \mathrm{x}\right)=0 \\
&\Rightarrow \sqrt{1-x^{2}}=0 \\
&\Rightarrow 1-\mathrm{x}^{2}=0 \\
&\Rightarrow \mathrm{x}=\pm 1
\end{aligned}
$$
$\mathrm{x}=-1$ not possible $\cos ^{-1}(-1)+\sin ^{-1}\left(-\frac{1}{2}\right)=\pi+\frac{5 \pi}{5} \neq \frac{\pi}{6}$
$$
\therefore \mathrm{x}=1
$$

22) Answer : (d)
Explanation:
$$
\cos ^{-1}\left(\cos \left(-\frac{\pi}{3}\right)\right)=\cos ^{-1}\left(\cos \left(\frac{\pi}{3}\right)\right)=\frac{\pi}{3}
$$
because $\cos \theta$ is positive in the fourth quadrant.

23) Answer : (c)
Explanation:
$$
\begin{aligned}
&=\tan ^{-1}\left(\frac{1}{7}\right)+2 \tan ^{-1}\left(\frac{1}{3}\right) \\
&=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{2 \cdot \frac{1}{3}}{1-\left(\frac{1}{3}\right)^{2}}\right) \\
&=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{3}{4}\right) \\
&=\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7} \cdot \frac{3}{4}}\right) \\
&=\tan ^{-1}(1)=\frac{\pi}{4}
\end{aligned}
$$

24) Answer: (d)
Explanation:
$$
\begin{aligned}
&\text { Put } \sec ^{-1} 2=A, \operatorname{cosec}^{-1} 3=B \\
&=\tan ^{2}\left(\sec ^{-1} 2\right)+\cot ^{2}\left(\operatorname{cosec}^{-1} 3\right) \\
&=\tan ^{2}(A)+\cot ^{2}(B) \\
&=\sec ^{2} A-1+\operatorname{cosec}^{2} B-1 \\
&=\left[\sec \left(\sec ^{-1} 2\right)\right]^{2}+\left[\operatorname{cosec}\left(\operatorname{cosec}^{-1} 3\right)\right]^{2}-2 \\
&=4+9-2 \\
&=11
\end{aligned}
$$

25) Answer : (d)
Explanation:
$$
\begin{aligned}
&\cot ^{-1}(\sqrt{\cos \alpha})+\tan ^{-1}(\sqrt{\cos \alpha})=\mu \\
&\operatorname{let} \sqrt{\cos \cos \alpha}=\theta \\
&\cot ^{-1} \theta+\tan ^{-1} \theta=\mu \\
&\Rightarrow \frac{\pi}{2}=\mu \\
&\sin \mu=\sin \frac{\pi}{2}=1
\end{aligned}
$$

26) Answer : (c) 

Explanation:

As no value of x in (0, 1) can satisfy the given equation.

Thus, the given equation has only one solution.

27) Answer : (b)
Explanation:
$$
\begin{aligned}
&\tan ^{-1}\left(\frac{a}{b}\right)-\tan ^{-1}\left(\frac{a-b}{a+b}\right) \\
&=\tan ^{-1}\left(\frac{\frac{a}{b}-\frac{a-b}{a+b}}{1+\frac{a}{b} \cdot \frac{a-b}{a+b}}\right) \\
&=\tan ^{-1}\left(\frac{a^{2}+a b-a b+b^{2}}{a b+b^{2}+a^{2}-a b}\right) \\
&=\tan ^{-1}(1) \\
&=\frac{\pi}{4}
\end{aligned}
$$

28)  Answer : (c)

Explanation:

The relation is true for all real values of x greater than or equal to 1.

29) Answer : (c)
Explanation:
We know that,
$$
\begin{aligned}
\sin 2 \theta &=\frac{2 \tan \theta}{1+\tan ^{2} \theta} \\
\cos 2 \theta &=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta} \text { and } \\
\tan 2 \theta &=\frac{2 \tan \theta}{1-\tan ^{2} \theta}
\end{aligned}
$$

30) Answer: (a)
Explanation:
$$
\begin{aligned}
&\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right) \\
&=\tan ^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4} \cdot \frac{2}{9}}\right) \\
&=\tan ^{-1}\left(\frac{17}{34}\right) \\
&=\tan ^{-1}\left(\frac{1}{2}\right) \\
&=\frac{1}{2}\left(2 \tan ^{-1} \frac{1}{2}\right) \\
&=\frac{1}{2} \cos ^{-1}\left(\frac{1-\left(\frac{1}{2}\right)^{2}}{1+\left(\frac{1}{2}\right)^{2}}\right) \\
&=\frac{1}{2} \cos ^{-1}\left(\frac{3}{5}\right)
\end{aligned}
$$

31) Answer: (c)
Explanation:
Since $\tan ^{-1} \mathrm{x}$ is not defined for $\mathrm{x}=\frac{\pi}{2}$. so, the given equation has no solution.

32) Answer : (d)
Explanation:
$$
\begin{aligned}
&\cos ^{-1}(\sqrt{3} x)+\cos ^{-1} \mathrm{x}=\frac{\pi}{2} \\
&\Rightarrow \cos ^{-1}(\sqrt{3} x)=\frac{\pi}{2}-\cos ^{-1} \mathrm{x} \\
&\Rightarrow \cos ^{-1}(\sqrt{3} x)=\sin ^{-1} \mathrm{x} \\
&\Rightarrow \sqrt{3} \mathrm{x}=\cos \left(\sin ^{-1} \mathrm{x}\right)=\sqrt{1-x^{2}} \\
&\Rightarrow 3 \mathrm{x}^{2}=1-\mathrm{x}^{2} \\
&\Rightarrow \mathrm{x}=\pm \frac{1}{2} \\
&\mathrm{x}=-\frac{1}{2} \text { does not satisfy the given equation. } \\
&\therefore \mathrm{x}=+\frac{1}{2}
\end{aligned}
$$

33) Answer: (d)
Explanation:
$$
\begin{aligned}
&y=\sin ^{-1}\left(-x^{2}\right) \Rightarrow \operatorname{siny}=-x^{2} \\
&\text { i.e., }-1 \leq-x^{2} \leq 1 \\
&\Rightarrow 1 \geq x^{2} \geq 1 \\
&\Rightarrow 0 \leq x^{2} \leq 1 \\
&\Rightarrow|x| \leq 1 \text { i.e., }-1 \leq x \leq 1
\end{aligned}
$$

34) Answer: (c)
Explanation:
$$
\begin{aligned}
&\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4} \\
&\Rightarrow \tan ^{-1}\left[\frac{\left(\frac{x-1}{x-2}\right)+\left(\frac{x+1}{x+2}\right)}{1-\left(\frac{x-1}{x-2}\right)\left(\frac{x+1}{x+2}\right)}\right]=\frac{\pi}{4} \\
&\Rightarrow\left[\frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)-(x+1)(x-1)}\right]=1 \\
&\Rightarrow\left(\frac{2 x^{2}-4}{-3}\right)=1 \\
&\Rightarrow\left(\frac{2 x^{2}-4}{-3}\right)=1 \\
&\Rightarrow 2 x^{2}=1 \\
&\Rightarrow x=\pm \frac{1}{\sqrt{2}}
\end{aligned}
$$

35) Answer: $\quad$ (a)
Explanation:
Let $\sin ^{-1} \mathrm{x}=\theta$ then $\sin \theta=\mathrm{x}$
$$
\begin{aligned}
&\Rightarrow \operatorname{cosec} \theta=\frac{1}{x} \\
&\Longrightarrow \operatorname{cosec}^{2} \theta=\frac{1}{x^{2}} \\
&\Rightarrow 1+\cot ^{2} \theta=\frac{1}{x^{2}} \\
&\Rightarrow \cot \theta=\frac{\sqrt{1-x^{2}}}{x}
\end{aligned}
$$

36) Answer: (d)
Explanation:
$\left(\cos ^{-1} \mathrm{x}+\sin ^{-1} \mathrm{x}\right)=\frac{\pi}{2}$ then, $\mathrm{x}=\frac{1}{2}$
Because $\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$
$\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$

37) Answer: (c)
Explanation:
Given,
$$
\begin{aligned}
&\tan ^{-1} \mathrm{x}+\tan ^{-1}\left(\frac{1}{7}\right)=\frac{\pi}{4} \\
&\Rightarrow \tan ^{-1} \mathrm{x}=\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{7}\right) \\
&\Rightarrow \tan ^{-1} \mathrm{x}=\tan ^{-1} 1-\tan ^{-1}\left(\frac{1}{7}\right) \\
&\Rightarrow \tan ^{-1} \mathrm{x}=\tan ^{-1}\left(\frac{1-\frac{1}{7}}{1+\frac{1}{7}}\right)=\tan ^{-1}\left(\frac{3}{4}\right) \\
&\Rightarrow \tan ^{-1} \mathrm{x}=\tan ^{-1}\left(\frac{3}{4}\right) \\
&\therefore \mathrm{x}=\frac{3}{4}
\end{aligned}
$$

38) Answer: (a)
Explanation:
We know that,
$$
\begin{aligned}
&\tan ^{-1} \mathrm{x}+\cot ^{-1} \mathrm{x}=\frac{\pi}{2} \\
&\therefore \cot ^{-1} \mathrm{x}=\frac{\pi}{2}-\frac{\pi}{10}=\frac{2 \pi}{5}
\end{aligned}
$$

39) Answer: (c)
Explanation:
$$
\begin{aligned}
&\cot ^{-1}(21)+\cot ^{-1}(13)-\cot ^{-1}(8) \\
&\Rightarrow \tan ^{-1}\left(\frac{1}{21}\right)+\tan ^{-1}\left(\frac{1}{13}\right)-\tan ^{-1}\left(\frac{1}{8}\right) \\
&\Rightarrow \tan ^{-1}\left[\frac{\frac{1}{21}+\frac{1}{13}}{1-\frac{1}{21} \cdot \frac{1}{13}}\right]-\tan ^{-1}\left(\frac{1}{8}\right) \\
&\Rightarrow \tan ^{-1}\left(\frac{34}{272}\right)-\tan ^{-1}\left(\frac{1}{8}\right) \\
&\Rightarrow \tan ^{-1}\left(\frac{1}{8}\right)-\tan ^{-1}\left(\frac{1}{8}\right)=0
\end{aligned}
$$

40) Answer: (d)
Explanation:
$$
\begin{aligned}
&\text { Let } \cos ^{-1}(-1)=\mathrm{A} \\
&\Rightarrow \cos \mathrm{A}=-1 \\
&\Rightarrow \cos \mathrm{A}=\cos \pi \\
&\Rightarrow \mathrm{A}=\pi \\
&\sin ^{-1}(1)=\mathrm{B} \\
&\Rightarrow \sin \mathrm{B}=1 \\
&\Rightarrow \sin \mathrm{B}=\sin \frac{\pi}{2} \\
&\rightarrow \mathrm{B}=\frac{\pi}{2} \\
&\cos ^{-1}(-1)-\sin ^{-1}(1)=\frac{\pi}{2}-\pi=\frac{\pi}{2}
\end{aligned}
$$

41) Answer: (c)
Explanation:
$$
\begin{aligned}
&\cot ^{-1} \mathrm{x}=\theta \quad \Rightarrow \mathrm{x}=\cot \theta \\
&\Rightarrow \cot \theta=\frac{x}{1} \\
&\sin \left(\cot ^{-1} \mathrm{x}\right)=\sin \theta=\frac{1}{\sqrt{1+x^{2}}}
\end{aligned}
$$

42) Answer: (a)
Explanation:
$$
\begin{aligned}
\sin ^{-1}\left(\sin \frac{3 \pi}{4}\right)=\sin ^{-1}\left(\sin \left(\pi-\frac{\pi}{4}\right)\right) \\
&=\sin ^{-1}\left(\sin \left(\frac{\pi}{4}\right)\right)=\frac{\pi}{4}
\end{aligned}
$$

43) Answer : (d)
Explanation:
$$
\begin{aligned}
&y=\cos ^{-1}\left(x^{2}-4\right) \\
&\Rightarrow \cos y=x^{2}-4 \\
&\text { i.e., }-1 \leq x^{2}-4 \leq 1
\end{aligned}
$$
now, $x^{2} \leq 5 \Rightarrow \sqrt{5} \leq|x| \leq \sqrt{5}$
and $3 \leq \mathrm{x} \Rightarrow \mathrm{x} \leq-\sqrt{3}$ and $\mathrm{x} \geq \sqrt{3}$
$\Rightarrow \mathrm{x} \in[-\sqrt{5},-\sqrt{3}] \cup[\sqrt{3}, \sqrt{5}]$

44) Answer : (a)
Explanation:
$$
\begin{aligned}
&=\sin ^{-1}\left(\cos \left(\frac{43 \pi}{5}\right)\right) \\
&=\sin ^{-1}\left(\cos \left(\frac{40 \pi+30 \pi}{5}\right)\right) \\
&=\sin ^{-1}\left(\cos \left(8 \pi+\frac{3 \pi}{5}\right)\right) \\
&=\sin ^{-1}\left(\cos \left(\frac{3 \pi}{5}\right)\right) \\
&=\sin ^{-1}\left(\sin \left(\frac{\pi}{2}-\frac{3 \pi}{5}\right)\right) \\
&=\sin ^{-1}\left(\sin \left(-\frac{\pi}{10}\right)\right) \\
&=-\frac{\pi}{10} .
\end{aligned}
$$

45) Answer: (a)
Explanation:
$$
\begin{aligned}
&\left(\sin ^{-1} x\right)^{2}+\left(\cos ^{-1} x\right)^{2}=\left(\sin ^{-1} x+\cos ^{-1} x\right)^{2}-2 \sin ^{-1} x \cos ^{-1} x \\
&=\frac{\pi^{2}}{4}-2 \sin ^{-1} x\left(\frac{\pi}{2}-\sin ^{-1} x\right) \\
&=\frac{\pi^{2}}{4}-\pi \sin ^{-1} x+2\left(\sin ^{-1} x\right)^{2} \\
&=2\left[\left(\sin ^{-1} x\right)^{2}-\frac{\pi}{2} \sin ^{-1} x+\frac{\pi^{2}}{8}\right] \\
&=2\left[\left(\sin ^{-1} x-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{16}\right]
\end{aligned}
$$
Thus, the least value of $2 \frac{\pi^{2}}{16}$ i.e. $\frac{\pi^{2}}{8}$ and the greatest value is $2\left[\left(\frac{-\pi}{2}-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{16}\right]$ i.e., $\frac{5 \pi^{2}}{4}$

46) Answer: (d)
Explanation:
$$
\begin{aligned}
&\cos ^{-1}\left[\cos \left(-680^{\circ}\right)\right] \\
&=\cos ^{-1}\left[\cos \left(720^{\circ}-40^{\circ}\right)\right] \\
&=\cos ^{-1}\left[\cos \left(-40^{\circ}\right)\right] \\
&=\cos ^{-1}\left[\cos \left(40^{\circ}\right)\right] \\
&=40^{\circ} \\
&=\frac{2 \pi}{9}
\end{aligned}
$$ 

47) Answer: (b)
Explanation:
$$
\begin{aligned}
&\cos ^{-1} \mathrm{x}=\theta \Rightarrow \mathrm{x}=\cos \theta \\
&\Rightarrow \cos \theta=\frac{x}{1} \\
&\cot \left(\cos ^{-1} \mathrm{x}\right)=\cot \theta=\frac{x}{\sqrt{1-x^{2}}}
\end{aligned}
$$

48) Answer: (a)
Explanation:
$$
[2 \pi, 3 \pi]
$$

49) Answer: (a)
Explanation:
$\sin ^{-1}(0.6)=\theta$ i.e., $\sin \theta=0.6$
Now, $\sin 2 \theta=2 \sin \theta \cos \theta$
$$
=2(0.6)(0.8)
$$
$=0.96$

50) Answer : (b)
Explanation:
$$
\cos ^{-1}(\cos \mathrm{x})=\mathrm{x} \text { if } 0 \leq \mathrm{x} \leq \pi \text { i.e., } \mathrm{x} \in[0, \pi]
$$

CASE STUDY QUESTIONS - SOLUTIONS

QUESTION - 1

1 (b)

2 (c)

3 ( d )

4 (c)

5 (c )

QUESTION - 2 ANSWERS
1. (b) $\tan ^{-1}\left(\frac{1}{2}\right)$
2. (c) $\tan ^{-1}\left(\frac{4}{3}\right)$
3. (d) $\tan ^{-1}\left(\frac{11}{2}\right)$
4. (b) $\tan ^{-1}(1 / 8)$
5. (c) $R,\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$