Q: Prove $3 \cos ^{-1} x=\cos ^{-1}\left(4 x^{3}-3 x\right), x \in\left[\frac{1}{2}, 1\right]$.

Let $x=\theta$ os Hence, $\cos ^{-1}(x)=$ Now, $$ \begin{aligned} R H S & =\cos ^{-1}\left(4 x^{3}-3 x\right) \\ & =\theta \cos 3^{1}\left(\cos \theta \cos ^{3}-\right. \\ & =\theta \cos ^{-1}(\cos 3) \\ & =\theta \\ & =3 \cos ^{-1} x \\ & =L H S \end{aligned} $$

Q: Prove $\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}$.

Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}$ Now, $$ \begin{aligned} L H S & =\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24} \\ & =\tan ^{-1} \frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11} \cdot \frac{7}{24}} \\ & =\tan ^{-1}\left(\frac{\frac{48+77}{264}}{\frac{264-14}{264}}\right) \\ & =\tan ^{-1}\left(\frac{125}{250}\right) \\ & =\tan ^{-1}\left(\frac{1}{2}\right) \\ & =\text { RHS } \end{aligned} $$

Q: Prove $2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}$.

Since we know that $2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}$ and $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}$ Now, $$ \begin{aligned} L H S & =2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7} \\ & =\tan ^{-1} \frac{2 \times \frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}+\tan ^{-1} \frac{1}{7} \\ & =\tan ^{-1}\left(\frac{4}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right) \\ & =\tan ^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3} \cdot \frac{1}{7}}\right) \\ & =\tan ^{-1}\left(\frac{\frac{28+3}{21}}{\frac{21-4}{21}}\right) \\ & =\tan ^{-1}\left(\frac{31}{17}\right) \\ & =R H S \end{aligned} $$

Q: Write the function in the simplest form: $\tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}, x \neq 0$

Let $x=\tan \theta \Rightarrow \theta=\tan ^{-1} x$ Hence, $$ \begin{aligned} \tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x} & =\tan ^{-1}\left(\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right) \\ & =\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right) \\ & =\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) \\ & =\tan ^{-1}\left(\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right) \\ & =\tan ^{-1}\left(\tan \frac{\theta}{2}\right) \\ & =\frac{\theta}{2} \\ & =\frac{1}{2} \tan ^{-1} x \end{aligned} $$

Q: Write the function in the simplest form: $\tan ^{-1} \frac{1}{\sqrt{x^{2}-1}},|x|>1$

Let $x=\operatorname{cosec} \theta \Rightarrow \theta=\operatorname{cosec}^{-1} x$ Hence, $$ \begin{aligned} \tan ^{-1} \frac{1}{\sqrt{x^{2}-1}} & =\tan ^{-1} \frac{1}{\sqrt{\operatorname{cosec}^{2} \theta-1}} \\ & =\tan ^{-1}\left(\frac{1}{\cot \theta}\right) \\ & =\tan ^{-1}(\tan \theta) \\ & =\theta \\ & =\operatorname{cosec}^{-1} x \\ & =\frac{\pi}{2}-\sec ^{-1} x \end{aligned} $$

Q: Write the function in the simplest form: $\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right), 0<x<\pi$

Since, $1-\cos x=2 \sin ^{2} \frac{x}{2}$ and $1+\cos x=2 \cos ^{2} \frac{x}{2}$ Hence, $$ \begin{aligned} \tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right) & =\tan ^{-1}\left(\sqrt{\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}}\right) \\ & =\tan ^{-1}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right) \\ & =\tan ^{-1}\left(\tan \frac{x}{2}\right) \\ & =\frac{x}{2} \end{aligned} $$

Q: Write the function in the simplest form: $\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right), 0<x<\pi$

$$ \begin{aligned} \tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right) & =\tan ^{-1}\left(\frac{\frac{\cos x-\sin x}{\cos x}}{\frac{\cos x+\sin x}{\cos x}}\right) \\ & =\tan ^{-1}\left(\frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}\right) \\ & =\tan ^{-1}\left(\frac{1-\tan x}{1+\tan x}\right) \\ & =\tan ^{-1}(1)-\tan ^{-1}(\tan x) \\ & =\frac{\pi}{4}-x \end{aligned} $$

Q: Write the function in the simplest form: $\tan ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}},|x|<a$

Let $x=a \sin \theta \Rightarrow \theta=\sin ^{-1}\left(\frac{x}{a}\right)$ Hence, $$ \begin{aligned} \tan ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}} & =\tan ^{-1}\left(\frac{a \sin \theta}{\sqrt{a^{2}-a^{2} \sin ^{2} \theta}}\right) \\ & =\tan ^{-1}\left(\frac{a \sin \theta}{a \sqrt{1-\sin ^{2} \theta}}\right) \\ & =\tan ^{-1}\left(\frac{a \sin \theta}{a \cos \theta}\right) \\ & =\tan ^{-1}(\tan \theta) \\ & =\theta \\ & =\sin ^{-1} \frac{x}{a} \end{aligned} $$

Q: Write the function in the simplest form: $\tan ^{-1}\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right), a>0 ; \frac{-a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}$

Let $x=a \tan \theta \Rightarrow \theta=\tan ^{-1}\left(\frac{x}{a}\right)$ Hence, $$ \begin{aligned} \tan ^{-1}\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right) & =\tan ^{-1}\left(\frac{3 a^{2} \cdot a \tan \theta-a^{3} \tan ^{3} \theta}{a^{3}-3 a \cdot a^{2} \tan ^{2} \theta}\right) \\ & =\tan ^{-1}\left(\frac{3 a^{3} \tan \theta-a^{3} \tan ^{3} \theta}{a^{3}-3 a^{3} \tan ^{2} \theta}\right) \\ & =\tan ^{-1}(\tan 3 \theta) \\ & =3 \theta \\ & =3 \tan ^{-1} \frac{x}{a} \end{aligned} $$

Question 1

Prove $3 \sin ^{-1} x=\sin ^{-1}\left(3 x-4 x^{3}\right), x \in\left[-\frac{1}{2}, \frac{1}{2}\right]$.

Accepted Answer

Let $x=\operatorname{Sin}$

Hence, $\sin ^{-1}(x)=$

Now,

$$

\begin{aligned}

R H S & =\sin ^{-1}\left(3 x-4 x^{3}ight) \

& =\sin 4(\sin \sin -\quad 3) \

& =\sin ^{-1}(\sin 3) \

& =	heta \

& =3 \sin ^{-1} x \

& =L H S

\end{aligned}

$$

Question 2

Prove $3 \cos ^{-1} x=\cos ^{-1}\left(4 x^{3}-3 x\right), x \in\left[\frac{1}{2}, 1\right]$.

Accepted Answer

Let $x=	heta$ os

Hence, $\cos ^{-1}(x)=$

Now,

$$

\begin{aligned}

R H S & =\cos ^{-1}\left(4 x^{3}-3 xight) \

& =	heta \cos 3^{1}\left(\cos 	heta \cos ^{3}-ight. \

& =	heta \cos ^{-1}(\cos 3) \

& =	heta \

& =3 \cos ^{-1} x \

& =L H S

\end{aligned}

$$

Question 3

Prove $	an ^{-1} \frac{2}{11}+	an ^{-1} \frac{7}{24}=	an ^{-1} \frac{1}{2}$.

Accepted Answer

Since we know that $	an ^{-1} x+	an ^{-1} y=	an ^{-1} \frac{x+y}{1-x y}$

Now,

$$

\begin{aligned}

L H S & =	an ^{-1} \frac{2}{11}+	an ^{-1} \frac{7}{24} \

& =	an ^{-1} \frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11} \cdot \frac{7}{24}} \

& =	an ^{-1}\left(\frac{\frac{48+77}{264}}{\frac{264-14}{264}}ight) \

& =	an ^{-1}\left(\frac{125}{250}ight) \

& =	an ^{-1}\left(\frac{1}{2}ight) \

& =	ext { RHS }

\end{aligned}

$$

Question 4

Prove $2 	an ^{-1} \frac{1}{2}+	an ^{-1} \frac{1}{7}=	an ^{-1} \frac{31}{17}$.

Accepted Answer

Since we know that $2 	an ^{-1} x=	an ^{-1} \frac{2 x}{1-x^{2}}$ and $	an ^{-1} x+	an ^{-1} y=	an ^{-1} \frac{x+y}{1-x y}$ Now,

$$

\begin{aligned}

L H S & =2 	an ^{-1} \frac{1}{2}+	an ^{-1} \frac{1}{7} \

& =	an ^{-1} \frac{2 	imes \frac{1}{2}}{1-\left(\frac{1}{2}ight)^{2}}+	an ^{-1} \frac{1}{7} \

& =	an ^{-1}\left(\frac{4}{3}ight)+	an ^{-1}\left(\frac{1}{7}ight) \

& =	an ^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3} \cdot \frac{1}{7}}ight) \

& =	an ^{-1}\left(\frac{\frac{28+3}{21}}{\frac{21-4}{21}}ight) \

& =	an ^{-1}\left(\frac{31}{17}ight) \

& =R H S

\end{aligned}

$$

Question 5

Write the function in the simplest form: $	an ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}, x 
eq 0$

Accepted Answer

Let $x=	an 	heta \Rightarrow 	heta=	an ^{-1} x$

Hence,

$$

\begin{aligned}

an ^{-1} \frac{\sqrt{1+x^{2}}-1}{x} & =	an ^{-1}\left(\frac{\sqrt{1+	an ^{2} 	heta}-1}{	an 	heta}ight) \

& =	an ^{-1}\left(\frac{\sec 	heta-1}{	an 	heta}ight) \

& =	an ^{-1}\left(\frac{1-\cos 	heta}{\sin 	heta}ight) \

& =	an ^{-1}\left(\frac{2 \sin ^{2} \frac{	heta}{2}}{2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}}ight) \

& =	an ^{-1}\left(	an \frac{	heta}{2}ight) \

& =\frac{	heta}{2} \

& =\frac{1}{2} 	an ^{-1} x

\end{aligned}

$$

Question 6

Write the function in the simplest form: $	an ^{-1} \frac{1}{\sqrt{x^{2}-1}},|x|>1$

Accepted Answer

Let $x=\operatorname{cosec} 	heta \Rightarrow 	heta=\operatorname{cosec}^{-1} x$

Hence,

$$

\begin{aligned}

an ^{-1} \frac{1}{\sqrt{x^{2}-1}} & =	an ^{-1} \frac{1}{\sqrt{\operatorname{cosec}^{2} 	heta-1}} \

& =	an ^{-1}\left(\frac{1}{\cot 	heta}ight) \

& =	an ^{-1}(	an 	heta) \

& =	heta \

& =\operatorname{cosec}^{-1} x \

& =\frac{\pi}{2}-\sec ^{-1} x

\end{aligned}

$$

Question 7

Write the function in the simplest form: $	an ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}ight), 0<x<\pi$

Accepted Answer

Since, $1-\cos x=2 \sin ^{2} \frac{x}{2}$ and $1+\cos x=2 \cos ^{2} \frac{x}{2}$

Hence,

$$

\begin{aligned}

an ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}ight) & =	an ^{-1}\left(\sqrt{\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}}ight) \

& =	an ^{-1}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}ight) \

& =	an ^{-1}\left(	an \frac{x}{2}ight) \

& =\frac{x}{2}

\end{aligned}

$$

Question 8

Write the function in the simplest form: $	an ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}ight), 0<x<\pi$

Accepted Answer

$$

\begin{aligned}

an ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}ight) & =	an ^{-1}\left(\frac{\frac{\cos x-\sin x}{\cos x}}{\frac{\cos x+\sin x}{\cos x}}ight) \

& =	an ^{-1}\left(\frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}ight) \

& =	an ^{-1}\left(\frac{1-	an x}{1+	an x}ight) \

& =	an ^{-1}(1)-	an ^{-1}(	an x) \

& =\frac{\pi}{4}-x

\end{aligned}

$$

Question 9

Write the function in the simplest form: $	an ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}},|x|<a$

Accepted Answer

Let $x=a \sin 	heta \Rightarrow 	heta=\sin ^{-1}\left(\frac{x}{a}ight)$

Hence,

$$

\begin{aligned}

an ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}} & =	an ^{-1}\left(\frac{a \sin 	heta}{\sqrt{a^{2}-a^{2} \sin ^{2} 	heta}}ight) \

& =	an ^{-1}\left(\frac{a \sin 	heta}{a \sqrt{1-\sin ^{2} 	heta}}ight) \

& =	an ^{-1}\left(\frac{a \sin 	heta}{a \cos 	heta}ight) \

& =	an ^{-1}(	an 	heta) \

& =	heta \

& =\sin ^{-1} \frac{x}{a}

\end{aligned}

$$

Question 10

Write the function in the simplest form: $	an ^{-1}\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}ight), a>0 ; \frac{-a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}$

Accepted Answer

Let $x=a 	an 	heta \Rightarrow 	heta=	an ^{-1}\left(\frac{x}{a}ight)$

Hence,

$$

\begin{aligned}

an ^{-1}\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}ight) & =	an ^{-1}\left(\frac{3 a^{2} \cdot a 	an 	heta-a^{3} 	an ^{3} 	heta}{a^{3}-3 a \cdot a^{2} 	an ^{2} 	heta}ight) \

& =	an ^{-1}\left(\frac{3 a^{3} 	an 	heta-a^{3} 	an ^{3} 	heta}{a^{3}-3 a^{3} 	an ^{2} 	heta}ight) \

& =	an ^{-1}(	an 3 	heta) \

& =3 	heta \

& =3 	an ^{-1} \frac{x}{a}

\end{aligned}

$$

Question 11

Write the function in the simplest form: $	an ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}ight)ight]$

Accepted Answer

Let $\sin ^{-1} \frac{1}{2}=x$

Hence,

$$

\begin{aligned}

\sin x & =\frac{1}{2} \

& =\sin \left(\frac{\pi}{6}ight) \

x & =\frac{\pi}{6} \

\sin ^{-1}\left(\frac{1}{2}ight) & =\frac{\pi}{6}

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

an ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}ight)ight] & =	an ^{-1}\left[2 \cos \left(2 	imes \frac{\pi}{6}ight)ight] \

& =	an ^{-1}\left[2 \cos \frac{\pi}{3}ight] \

& =	an ^{-1}\left[2 	imes \frac{1}{2}ight] \

& =	an ^{-1}[1] \

& =\frac{\pi}{4}

\end{aligned}

$$

Question 12

Find the value of $\cot \left(	an ^{-1} a+\cot ^{-1} aight)$

Accepted Answer

Since $	an ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$

Hence,

$$

\begin{aligned}

\cot \left(	an ^{-1} a+\cot ^{-1} aight) & =\cot \left(\frac{\pi}{2}ight) \

& =0

\end{aligned}

$$

Question 13

Find the value of $	an \frac{1}{2}\left(\sin ^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}ight),|x|<1, y>0$ and $x y<1$.

Accepted Answer

Let $x=	an 	heta \Rightarrow 	heta=	an ^{-1} x$

Hence,

$$

\begin{aligned}

\sin ^{-1} \frac{2 x}{1+x^{2}} & =\sin ^{-1}\left(\frac{2 	an 	heta}{1+	an ^{2} 	heta}ight) \

& =\sin ^{-1}(\sin 2 	heta) \

& =2 	heta \

& =2 	an ^{-1} x

\end{aligned}

$$

Now, let $y=	an \phi \Rightarrow \phi=	an ^{-1} y$

Hence,

$$

\begin{aligned}

\cos ^{-1} \frac{1-y^{2}}{1+y^{2}} & =\cos ^{-1}\left(\frac{1-	an ^{2} \phi}{1+	an ^{2} \phi}ight) \

& =\cos ^{-1}(\cos 2 \phi) \

& =2 \phi \

& =2 	an ^{-1} y

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

an \frac{1}{2}\left(\sin ^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}ight) & =	an \frac{1}{2}\left(2 	an ^{-1} x+2 	an ^{-1} yight) \

& =	an \left(	an ^{-1} x+	an ^{-1} yight) \

& =	an \left[	an ^{-1}\left(\frac{x+y}{1-x y}ight)ight] \

& =\left(\frac{x+y}{1-x y}ight)

\end{aligned}

$$

Question 14

If $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$, find the value of $x$.

Accepted Answer

It is given that $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} xight)=1$

Since we know that $\sin (x+y)=\sin x \cos y+\cos x \sin y$ Therefore,

$$

\begin{align*}

\sin \left(\sin ^{-1} \frac{1}{5}ight) \cos \left(\cos ^{-1} xight)+\cos \left(\sin ^{-1} \frac{1}{5}ight) \sin \left(\cos ^{-1} xight) & =1 \

\left(\frac{1}{5}ight) 	imes(x)+\cos \left(\sin ^{-1} \frac{1}{5}ight) \sin \left(\cos ^{-1} xight) & =1 \

\frac{x}{5}+\cos \left(\sin ^{-1} \frac{1}{5}ight) \sin \left(\cos ^{-1} xight) & =1 	ag{1}

\end{align*}

$$

Now, let $\sin ^{-1} \frac{1}{5}=y \Rightarrow \sin y=\frac{1}{5}$

Then,

$$

\begin{aligned}

\cos y & =\sqrt{1-\left(\frac{1}{5}ight)^{2}} \

& =\frac{2 \sqrt{6}}{5} \

y & =\cos ^{-1}\left(\frac{2 \sqrt{6}}{5}ight)

\end{aligned}

$$

Therefore,

$$

\begin{equation*}

\sin ^{-1} \frac{1}{5}=\cos ^{-1}\left(\frac{2 \sqrt{6}}{5}ight) 	ag{2}

\end{equation*}

$$

Now, let $\cos ^{-1} x=z \Rightarrow \cos z=x$

Then,

$$

\begin{aligned}

\sin z & =\sqrt{1-x^{2}} \

z & =\sin ^{-1} \sqrt{1-x^{2}}

\end{aligned}

$$

Therefore,

$$

\begin{equation*}

\cos ^{-1} x=\sin ^{-1} \sqrt{1-x^{2}} 	ag{3}

\end{equation*}

$$

From (1),(2) and (3), we have

$$

\begin{aligned}

& \Rightarrow \frac{x}{5}+\cos \left(\cos ^{-1} \frac{2 \sqrt{6}}{5}ight) \sin \left(\sin ^{-1} \sqrt{1-x^{2}}ight)=1 \

& \Rightarrow \frac{x}{5}+\frac{2 \sqrt{6}}{5} \sqrt{1-x^{2}}=1 \

& \Rightarrow x+2 \sqrt{6} \sqrt{1-x^{2}}=5 \

& \Rightarrow 5-x=2 \sqrt{6} \sqrt{1-x^{2}}

\end{aligned}

$$

On squaring both the sides

$$

\begin{aligned}

25+x^{2}-10 x & =24-24 x^{2} \

25 x^{2}-10 x+1 & =0 \

(5 x-1)^{2} & =0 \

(5 x-1) & =0 \

x & =\frac{1}{5}

\end{aligned}

$$

Question 15

If $	an ^{-1} \frac{x-1}{x-2}+	an ^{-1} \frac{x+1}{x+2}=\frac{\pi}{4}$, find the value of $x$.

Accepted Answer

It is given that $	an ^{-1} \frac{x-1}{x-2}+	an ^{-1} \frac{x+1}{x+2}=\frac{\pi}{4}$

Since $	an ^{-1} x+	an ^{-1} y=	an ^{-1}\left(\frac{x+y}{1-x y}ight)$

Therefore,

$$

\begin{aligned}

& \Rightarrow 	an ^{-1}\left(\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\left(\frac{x-1}{x-2}ight)\left(\frac{x+1}{x+2}ight)}ight)=\frac{\pi}{4} \

& \Rightarrow 	an ^{-1}\left[\frac{(x-1)(x+2)+(x+1)(x-2)}{(x+2)(x-2)-(x-1)(x+1)}ight]=\frac{\pi}{4} \

& \Rightarrow 	an ^{-1}\left[\frac{x^{2}+x-2+x^{2}-x-2}{x^{2}-4-x^{2}+1}ight]=\frac{\pi}{4} \

& \Rightarrow 	an ^{-1}\left[\frac{2 x^{2}-4}{-3}ight]=\frac{\pi}{4} \

& \Rightarrow 	an \left[	an ^{-1} \frac{4-2 x^{2}}{3}ight]=	an \frac{\pi}{4} \

& \Rightarrow \frac{4-2 x^{2}}{3}=1 \

& \Rightarrow 4-2 x^{2}=3 \

& \Rightarrow 2 x^{2}=1 \

& \Rightarrow x^{2}=\frac{1}{2} \

& \Rightarrow x= \pm \frac{1}{\sqrt{2}}

\end{aligned}

$$

Question 16

Find the value of $\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$.

Accepted Answer

Since, ${ }^{\Theta}$ in $\sin (	heta-)$

Therefore,

$$

\begin{aligned}

\sin ^{-1}\left(\sin \frac{2 \pi}{3}ight) & =\sin ^{-1}\left[\sin \left(\pi-\frac{2 \pi}{3}ight)ight] \

& =\sin ^{-1}\left(\sin \frac{\pi}{3}ight) \

& =\frac{\pi}{3}

\end{aligned}

$$

Question 17

Find the value of $	an ^{-1}\left(	an \frac{3 \pi}{4}ight)$.

Accepted Answer

Since, tan tam ( $	heta-$ )

Therefore,

$$

\begin{aligned}

an ^{-1}\left(	an \frac{3 \pi}{4}ight) & =	an ^{-1}\left[-	an \left(-\frac{3 \pi}{4}ight)ight] \

& =	an ^{-1}\left[-	an \left(\pi-\frac{\pi}{4}ight)ight] \

& =	an ^{-1}\left[-	an \left(\frac{\pi}{4}ight)ight] \

& =	an ^{-1}\left[	an \left(-\frac{\pi}{4}ight)ight] \

& =\left(-\frac{\pi}{4}ight)

\end{aligned}

$$

Question 18

Find the value of $	an \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}ight)$.

Accepted Answer

Let $\sin ^{-1} \frac{3}{5}=x \Rightarrow \sin x=\frac{3}{5}$

Then,

$$

\begin{aligned}

& \Rightarrow \cos x=\sqrt{1-\sin ^{2} x}=\frac{4}{5} \

& \Rightarrow \sec x=\frac{5}{4}

\end{aligned}

$$

Therefore,

$$

\begin{align*}

an x & =\sqrt{\sec ^{2} x-1} \

& =\sqrt{\frac{25}{16}-1} \

& =\frac{3}{4} \

x & =	an ^{-1} \frac{3}{4} \

\sin ^{-1} \frac{3}{5} & =	an ^{-1} \frac{3}{4} 	ag{1}

\end{align*}

$$

Now,

$$

\begin{equation*}

\cot ^{-1} \frac{3}{2}=	an ^{-1} \frac{2}{3} 	ag{2}

\end{equation*}

$$

Thus, by using (1) and (2)

$$

\begin{aligned}

an \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{2}{3}ight) & =	an \left(	an ^{-1} \frac{3}{4}+	an ^{-1} \frac{2}{3}ight) \

& =	an \left[	an ^{-1} \frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}ight] \

& =	an \left(	an ^{-1} \frac{17}{6}ight) \

& =\frac{17}{6}

\end{aligned}

$$

Question 19

$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$ is equal to

(A) $\frac{7 \pi}{6}$

(B) $\frac{5 \pi}{6}$

(C) $\frac{\pi}{3}$

(D) $\frac{\pi}{6}$

Accepted Answer

$$

\begin{aligned}

\cos ^{-1}\left(\cos \frac{7 \pi}{6}ight) & =\cos ^{-1}\left(\cos \frac{-7 \pi}{6}ight) \

& =\cos ^{-1}\left[\cos \left(2 \pi-\frac{7 \pi}{6}ight)ight] \

& =\cos ^{-1}\left[\cos \left(\frac{5 \pi}{6}ight)ight] \

& =\frac{5 \pi}{6}

\end{aligned}

$$

Thus, the correct option is B.

Question 20

$\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)$ is equal to

(A) $\frac{1}{2}$

(B) $\frac{1}{3}$

(C) $\frac{1}{4}$

(D) 1

Accepted Answer

Let $\sin ^{-1}\left(-\frac{1}{2}ight)=x$

Hence,

$$

\begin{aligned}

\sin x & =-\frac{1}{2} \

& =-\sin \frac{\pi}{6} \

& =\sin \left(-\frac{\pi}{6}ight) \

x & =-\frac{\pi}{6}

\end{aligned}

$$

Since, Range of principal value of $\sin ^{-1}(x)=\left[\frac{-\pi}{2}, \frac{\pi}{2}ight]$.

Therefore,

$$

\sin ^{-1}\left(-\frac{1}{2}ight)=-\frac{\pi}{6}

$$

Then,

$$

\begin{aligned}

\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}ight)ight) & =\sin \left(\frac{\pi}{3}+\frac{\pi}{6}ight) \

& =\sin \left(\frac{\pi}{2}ight) \

& =1

\end{aligned}

$$

Thus, the correct option is D.

Question 21

Find the values of $	an ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3})$ is equal to

(A) $\pi$

(B) $-\frac{\pi}{2}$

(C) 0

(D) $2 \sqrt{3}$

Accepted Answer

Let $	an ^{-1} \sqrt{3}=x$

Hence,

$$

an x=\sqrt{3}=	an \frac{\pi}{3}, 	ext { where } \frac{\pi}{3} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}ight)

$$

Therefore, $	an ^{-1} \sqrt{3}=\frac{\pi}{3}$

Now, let $\cot ^{-1}(-\sqrt{3})=y$

Hence,

$$

\begin{aligned}

\cot y & =(-\sqrt{3}) \

& =-\cot \left(\frac{\pi}{6}ight) \

& =\cot \left(\pi-\frac{\pi}{6}ight) \

& =\cot \left(\frac{5 \pi}{6}ight)

\end{aligned}

$$

Since, Range of principal value of $\cot ^{-1} x=(0, \pi)$ Therefore,

$$

\cot ^{-1}(-\sqrt{3})=\frac{5 \pi}{6}

$$

Then,

$$

\begin{aligned}

an ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3}) & =\frac{\pi}{3}-\frac{5 \pi}{6} \

& =-\frac{\pi}{2}

\end{aligned}

$$

Thus, the correct option is B.

\section*{MISCELLANEOUS EXERCISE}