A $f$ unction $f: \mathrm{A} \rightarrow \mathrm{B}$ is invertible if $f$ it is a bijection. The inverse of $f$ is denoted by $f^{-1}$ and is defined as $f^{-1}(\mathrm{y})=\mathrm{x} \Leftrightarrow f(\mathrm{x})=\mathrm{y}$. Trigonometric functions are periodic and hence they are not bijective. But if we restrict their domains and codomains they can be made bijective and we can obtain their inverses.

1. $\operatorname{Sin}^{-1} \mathbf{x}$ :

The symbol $\sin ^{-1} \mathrm{x}$ or $\arcsin \mathrm{x}$ denote the angle $\theta$ so that $\sin \theta=\mathrm{x}$. As a direct meaning, $\sin ^{-1} \mathrm{x}$ is not a function, as it does not satisfy the requirements for a rule to become a function. But by a suitable choice $[-1,1]$ as its domain and standardized set $[-\pi / 2, \pi / 2]$ as its range, then rule $\sin ^{-}$ ${ }^{1} x$ is a single valued function

Thus $\sin ^{-1} \mathrm{x}$ is considered as a function with domain $[-1,1]$ and range $[-\pi / 2, \pi / 2]$

The graph of $y=\sin ^{-1} x$ is as shown below which is obtained by taking the mirror image, of the portion of the graph of $y=\sin x$ from $x=-\pi / 2$ to $x=\pi / 2$, on the line $y=x$

2. $\cos ^{-1} \mathbf{x}$ :

By following the discussions, similar to above, we have $\cos ^{-1} \mathrm{x}$ or arccos $\mathrm{x}$ as a function with domain $[-1,1]$ and range $[0, \pi]$

The graph of $y=\cos ^{-1} x$ is similarly obtained as the mirror image of the portion of the graph of $y=$ $\cos \mathrm{x}$ from $\mathrm{x}=0$ to $\mathrm{x}=\pi$

3. $\tan ^{-1} \mathbf{x}:$

We get $\tan ^{-1} x$ or arctan $x$ as a function with domain $R$ and range $(-\pi / 2, \pi / 2)$. Graph of $y=\tan ^{-1} x$

4. $\operatorname{cosec}^{-1} x:$

$\operatorname{cosec}^{-1} \mathrm{x}$ or arccosec $\mathrm{x}$ is a function with domain $\mathrm{R}-(-1,1)$ and range $[-\pi / 2, \pi / 2]-\{0\}$. Graph of $y=\operatorname{cosec}^{-1} \mathrm{x}$

5. $\sec ^{-1} \mathbf{x}$ :

$\sec ^{-1} x$ or arcsec $x$ is a function with domain $R-(-1,1)$ and range $[0, \pi]-\{\pi / 2\}$. Graph of $y=\sec -$ ${ }^{1} x$ is

6. $\cot ^{-1} \mathbf{x}$ :

$\cot ^{-1} \mathrm{x}$ or arccot $\mathrm{x}$ is a function with domain $\mathrm{R}$ and range $(0, \pi)$. Graph of $\mathrm{y}=\cot ^{-1} \mathrm{x}$ is

Property : $“-x"$

The graphs of $\sin ^{-1} x, \tan ^{-1} x, \operatorname{cosec}^{-1} x$ are symmetric about origin.

Hence we get

$ \begin{aligned} & \sin ^{-1}(-x)=-\sin ^{-1} x \\ & \tan ^{-1}(-x)=-\tan ^{-1} x \\ & \operatorname{cosec}^{-1}(-x)=-\operatorname{cosec}^{-1} x \end{aligned} $

Also the graphs of $\cos ^{-1} x, \sec ^{-1} x, \cot ^{-1} x$ are symmetric about the point $(0, \pi / 2)$. From this, we get

$ \begin{aligned} & \cos ^{-1}(-\mathrm{x})=\pi _{-}-\cos ^{-1} \mathrm{x} \\ & \sec ^{-1}(-\mathrm{x})=\pi _{-\sec ^{-1} \mathrm{x}} \\ & \cot ^{-1}(-\mathrm{x})=\pi _{-} \cot ^{-1} \mathrm{x} \end{aligned} $

Notes :

(i). $\mathrm{x}^{2}+\mathrm{y}^{2} \leq 1 \& \mathrm{x}, \mathrm{y} \geq 0 \Rightarrow \leq \sin ^{-1} \mathrm{x}+\sin ^{-1} \mathrm{y} \leq \frac{\pi}{2}$ and $\mathrm{x}^{2}+\mathrm{y}^{2} \geq 1$ \& $\mathrm{x}, \mathrm{y} \geq 0$

$ \Rightarrow \frac{\pi}{2} \leq \sin ^{-1} x+\sin ^{-1} y \leq \pi $

(ii). $x y<1$ and $x, y \geq 0 \Rightarrow 0 \leq \tan ^{-1} x+\tan ^{-1} y<\frac{\pi}{2}$; $x y>1$ and $x, y \geq 0 \Rightarrow \frac{\pi}{2}<\tan ^{-1} x+\tan ^{-1} y<\pi$

(iii). For $\mathrm{x}<0$ or $\mathrm{y}<0$ these identities can be used with the help of property " $-\mathrm{x}$ " i.e. change $x$ or $y$ to $-x$ or $-y$ which are positive

Domain & range of inverse trigonometric Functions

Note: If no branch of an inverse trigonometric function is mentioned, then it means the principal value branch of the function.

Properties of Inverse Trigonometric Functions

1. (i). $\sin ^{-1}(\sin \mathrm{x})=\left\{\begin{array}{l}-2 \mathrm{n} \pi+\mathrm{x}, \quad 2 \mathrm{n} \pi-\frac{\pi}{2} \leq \mathrm{x} \leq 2 \mathrm{n} \pi+\frac{\pi}{2}, \mathrm{n} \in \mathrm{Z} \\ (2 \mathrm{n}+1) \pi-\mathrm{x},(2 \mathrm{n}+1) \pi-\frac{\pi}{2} \leq \mathrm{x} \leq(2 \mathrm{n}+1) \pi+\frac{\pi}{2}, \mathrm{n} \in \mathrm{Z}\end{array}\right.$

Period $=2 \pi \&$ it is an odd function.

(ii). $\cos ^{-1}(\cos x)=\left\{\begin{array}{l}-2 \mathrm{n} \pi+x, \quad 2 \mathrm{n} \pi \leq \mathrm{x} \leq(2 \mathrm{n}+1) \pi, \quad \mathrm{n} \in \mathrm{Z} \\2\mathrm{n} \pi-x,(2 \mathrm{n}-1) \pi \leq \mathrm{x} \leq 2 \mathrm{n} \pi, \quad \mathrm{n} \in \mathrm{Z} \end{array}\right.$

Period $=2 \pi$ and it is an even function

(iii). $\tan ^{-1}(\tan \mathrm{x})=-\mathrm{n} \pi+\mathrm{x}, \mathrm{n} \pi-\frac{\pi}{2}<\mathrm{x}<\mathrm{n} \pi+\frac{\pi}{2}, \mathrm{n} \in \mathrm{Z}$

Period $=\pi$

(iv). $\cot ^{-1}(\operatorname{cotx})=-\mathrm{n} \pi+\mathrm{x}, \mathrm{n} \pi<\mathrm{x}<(\mathrm{n}+1) \pi, \mathrm{n} \in \mathrm{Z}$

Period $=\pi$

(v). $ \sec ^{-1}(\sec x)=\left\{\begin{array}{l}x-2 n \pi, 2 n \pi \leq x \leq(2 n+1) \pi, x \neq 2 \pi+\frac{\pi}{2} \\ -x+2 n \pi,(2 n-1) \pi \leq x \leq 2 n \pi, x \neq\left(2 n \pi-\frac{\pi}{2} , n \in Z\right.\end{array}\right.$

Period $=2 \pi$

(vi). $\operatorname{cosec}^{-1}(\operatorname{cosec} x)=\left\{\begin{array}{l}-2 n \pi+x, 2 n \pi-\frac{\pi}{2} \leq x \leq 2 n \pi+\frac{\pi}{2} \\ (2 n+1) \pi-x,(2 n+1) \pi-\frac{\pi}{2} \leq x \leq(2 n+1) \pi+\frac{\pi}{2}, x \neq n \pi, n \in Z\end{array}\right.$

Period $=2 \pi$

2. (i). $\sin \left(\sin ^{-1} x\right)=x,-1 \leq x \leq 1$ $\hspace{1cm}$(ii). $ \cos \left(\cos ^{-1} \mathrm{x}\right)=\mathrm{x},-1 \leq \mathrm{x} \leq 1$

(iii). $\tan \left(\tan ^{-1} x\right)=x, x \in R$ $\hspace{1cm}$(iv). $\cot \left(\cot ^{-1} \mathrm{x}\right)=\mathrm{x}, \mathrm{x} \in \mathrm{R}$

(v). $\sec \left(\sec ^{-1} x\right)=x, x \in R(-\infty,-1] \cup[1, \infty)$

(vi). $\operatorname{cosce}\left(\operatorname{cosce}^{-1} x\right)=x, x \in R(-\infty,-1] \cup[1, \infty)$

3. (i). $\sin ^{-1} x+\cos ^{-1} x=\pi / 2,-1 \leq x \leq 1$

(ii). $\tan ^{-1} x+\cot ^{-1} x=\pi / 2, x \in R$

(iii).$\sec ^{-1} \mathrm{x}+\operatorname{cosce}^{-1} \mathrm{x}=\pi / 2, \mathrm{x} \in \mathrm{R}(-\infty,-1] \cup[1, \infty)$

4. (i). $\sin ^{-1} x=\operatorname{cosec}^{-1} \frac{1}{x},-1 \leq x \leq 1$

(ii). $\cos ^{-1} x=\sec ^{-1}\left(\frac{1}{x}\right),-1 \leq x \leq 1$

(iii). $\tan ^{-1} x=\left\{\begin{array}{l}\cot ^{-1}(1 / x), x>0 \\ -\pi+\cot ^{-1}(1 / x), x<0\end{array}\right.$

5. (i). $\sin ^{-1}(-x)=-\sin ^{-1} x, \quad-1 \leq x \leq 1$

(ii). $\cos ^{-1}(-x)=\pi-\cos ^{-1} x, \quad-1 \leq x \leq 1$

(iii). $\tan ^{-1}(-x)=-\tan ^{-1} x, \quad-\infty<x<\infty$

(iv). $\cot ^{-1}(-x)=\pi-\cot ^{-1} x, \quad-\infty<x<\infty$

(v). $\operatorname{cosec}^{-1}(-x)=-\operatorname{cosec}^{-1} x, \quad x \leq-1$ or $x \geq 1$

(vi). $\sec ^{-1}(-x)=\pi-\sec ^{-1} x, \quad x \leq-1$ or $x \geq 1$

6. Conversions of inverse trigonometric functions

$ \begin{aligned} \sin ^{-1} x & =\cos ^{-1} \sqrt{1-x^{2}}=\tan ^{-1} \frac{x}{\sqrt{1-x^{2}}} \\ & =\cot ^{-1} \frac{\sqrt{1-x^{2}}}{x}=\operatorname{cosec}^{-1} \frac{1}{x}=\sec ^{-1} \frac{1}{\sqrt{1-x^{2}}} \end{aligned} $

$ \begin{aligned} \cos ^{-1} \mathrm{x} & =\sin ^{-1} \sqrt{1-\mathrm{x}^{2}}=\tan ^{-1} \frac{\sqrt{1-\mathrm{x}^{2}}}{\mathrm{x}} \\ & =\cot ^{-1} \frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}}=\operatorname{Cosec}^{-1} \frac{1}{\sqrt{1-\mathrm{x}^{2}}}=\sec ^{-1} \frac{1}{\mathrm{x}} \end{aligned} $

$ \begin{aligned} \tan ^{-1} \mathrm{x}= & \sin ^{-1} \frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}=\operatorname{Cos}^{-1} \frac{1}{\sqrt{1+\mathrm{x}^{2}}}=\cot ^{-1} \frac{1}{\mathrm{x}} \\ & \operatorname{cosec}^{-1} \frac{\sqrt{1+\mathrm{x}^{2}}}{\mathrm{x}}=\operatorname{Sec}^{-1} \sqrt{1+\mathrm{x}^{2}} \end{aligned} $

7. (i). $\sin ^{-1} x+\sin ^{-1} y=\left\{\begin{array}{l}\sin ^{-1}\left(x \sqrt{1-y^2}+y \sqrt{1-x^2}\right) \text { if }-1 \leq x, y \leq 1 &{\&} x^2+y^2 \leq 1 \\ \text { or if } x y<0 & {\&}x^2+y^2>1 \\ \pi-\sin ^{-1}\left(x \sqrt{1-y^2}+y \sqrt{1-x^2}\right) \text { if } 0<x, y \leq 1 & {\&}x^2+y^2>1 \\ -\pi-\sin ^{-1}\left(x \sqrt{1-y^2}+y \sqrt{1-x^2}\right) \text { if }-1 \leq x, y<0 & {\&}x^2+y^2>1\end{array}\right.$

(ii). $\sin ^{-1} x-\sin ^{-1} y=\left\{\begin{array}{l}\sin ^{-1}\left(x \sqrt{1-y^2}-y \sqrt{1-x^2}\right) \text { if }-1 \leq x, y \leq 1 & x^2+y^2 \leq 1 \\ \text { or if } x y>0 & x^2+y^2>1 \\ \pi-\sin ^{-1}\left(x \sqrt{1-y^2}-y \sqrt{1-x^2}\right) \text { if } 0<x \leq 1,-1<y \leq 0 & x^2+y^2>1 \\ -\pi-\sin ^{-1}\left(x \sqrt{1-y^2}-y \sqrt{1-x^2}\right) \text { if }-1 \leq x<0,0<y \leq 1 & x^2+y^2>1\end{array}\right.$

8. (i). $\cos ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{y}=\left\{\begin{array}{l}\cos ^{-1}\left(\mathrm{xy}-\sqrt{1-\mathrm{x}^{2}} \sqrt{1-\mathrm{y}^{2}}\right) \text { if }-1 \leq \mathrm{x}, \mathrm{y} \leq 1 \& \mathrm{x}+\mathrm{y} \geq 0 \\ 2 \pi-\cos ^{-1}\left(\mathrm{xy}-\sqrt{1-\mathrm{x}^{2}} \sqrt{1-\mathrm{y}^{2}}\right) \text { if }-1 \leq \mathrm{x}, \mathrm{y} \leq 1 \& \mathrm{x}+\mathrm{y} \leq 0\end{array}\right.$

(ii). $\quad \cos ^{-1} x-\cos ^{-1} y=\left\{\begin{array}{l}\cos ^{-1}\left(x y+\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right) \text { if }-1 \leq x, y \leq 1 \& x \leq y \\ -\cos ^{-1}\left(x y+\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right) \text { if }-1 \leq y \leq 0,0<x \leq 1 \& x \geq y\end{array}\right.$

9. (i). $\tan ^{-1} x+\tan ^{-1} y=\left\{\begin{array}{l}\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \text { if } x y<1 \\ \pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \text { if } x>0, y>0 x y>1 \\ -\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \text { if } x<0, y<0 x y>1\end{array}\right.$

(ii). $\tan ^{-1} x-\tan ^{-1} y=\left\{\begin{array}{l}\tan ^{-1}\left(\frac{x-y}{1+x y}\right) \text { if } x y>-1 \\ \pi+\tan ^{-1}\left(\frac{x-y}{1+x y}\right) \text { if } x>0, y<0 \& x y<-1 \\ -\pi+\tan ^{-1}\left(\frac{x-y}{1+x y}\right), \text { if } x<0, y>0 \text { \& } x y<-1\end{array}\right.$

Remark : If $x _{1}, \quad x _{2}, \ldots \ldots \ldots . x _{n} \in R$, then $\tan ^{-1} x _{1}+\tan ^{-1} x _{2}+\ldots \ldots \ldots+\tan ^{-1} x _{n}$ $=\tan ^{-1}\left(\frac{\mathrm{s} _{1}-\mathrm{s} _{3}+\mathrm{s} _{5}-\mathrm{s} _{7} \ldots \ldots . .}{1-\mathrm{s} _{2}+\mathrm{s} _{4}-\mathrm{s} _{6}+\ldots \ldots .}\right)$

Where $\mathrm{s} _{\mathrm{k}}$ is the sum of the product of $\mathrm{x} _{1}, \mathrm{x} _{2}……….x_n$ taken $k$ at $a$ time.

ie. $\mathrm{s} _{1}=\mathrm{x} _{1}+\mathrm{x} _{2}+\ldots \ldots \ldots+\mathrm{x} _{\mathrm{n}}=\sum \mathrm{x} _{\mathrm{i}}$

$ \mathrm{s} _{2}=\quad \mathrm{x} _{1} \mathrm{x} _{2}+\mathrm{x} _{2} \mathrm{x} _{3}+\ldots \ldots .+\mathrm{x} _{\mathrm{n}-1}^{\mathrm{n}} \mathrm{x} _{\mathrm{n}}=\sum \mathrm{x} _{1} \mathrm{x} _{2} $

$\mathrm{s} _{3}=\quad \sum \mathrm{x} _{1} \mathrm{x} _{2} \mathrm{x} _{3}………$etc.

10. (i). $\quad 2 \sin ^{-1} x=\left\{\begin{array}{l}\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right) \text { if } \frac{-1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}} \\ \pi-\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right), \text { if } \frac{1}{\sqrt{2}} \leq x \leq 1 \\ -\pi-\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right), \text { if }-1 \leq x \leq \frac{-1}{\sqrt{2}}\end{array}\right.$

(ii). $3 \sin ^{-1} x=\left\{\begin{array}{l}\sin ^{-1}\left(3 x-4 x^{3}\right), \text { if } \frac{-1}{2} \leq x \leq \frac{1}{2} \\ \pi-\sin ^{-1}\left(3 x-4 x^{3}\right), \text { if } \frac{1}{2}<x \leq 1 \\ -\pi-\sin ^{-1}\left(3 x-4 x^{3}\right), \text { if }-1 \leq x<\frac{-1}{2}\end{array}\right.$

11. (i). $2 \cos ^{-1} x=\left\{\begin{array}{l}\cos ^{-1}\left(2 x^{2}-1\right), \text { if } 0 \leq x \leq 1 \\ 2 \pi-\cos ^{-1}\left(2 x^{2}-1\right) \text { if }-1 \leq x \leq 0\end{array}\right.$

(ii). $\quad 3 \cos ^{-1} x=\left\{\begin{array}{l}\cos ^{-1}\left(4 x^{3}-3 x\right), \text { if } \frac{1}{2} \leq x \leq 1 \\ 2 \pi-\cos ^{-1}\left(4 x^{3}-3 x\right), \text { if } \frac{-1}{2} \leq x \leq \frac{1}{2} \\ 2 \pi+\cos ^{-1}\left(4 x^{3}-3 x\right), \text { if }-1 \leq x \leq \frac{-1}{2}\end{array}\right.$

12. (i). $2 \tan ^{-1} x=\left\{\begin{array}{l}\tan ^{-1}\left(\frac{2 x}{1-x^2}\right), \text { if }-1<x<1 \\ \pi+\tan ^{-1}\left(\frac{2 x}{1-x^2}\right), \text { if } x>1 \\ -\pi+\tan ^{-1}\left(\frac{2 x}{1-x^2}\right), \text { if } x<-1\end{array}\right.$

(ii). $3 \tan ^{-1} x=\left\{\begin{array}{l}\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right), \text { if } \frac{-1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}} \\ \pi+\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right), \text { if } x>\frac{1}{\sqrt{3}} \\ -\pi+\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right), \text { if } x<\frac{1}{\sqrt{3}}\end{array}\right.$

Note: If $|x| \leq 1$ then $2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$.

If $|x|>1$, change $x$ to $\frac{1}{x}$ in the above.

Note: In cases of identities in inverse trigonometric functions, principal values are to be taken. As such signs of $x, y$ etc., will determine the quadrant in which the angles will fall. In order to bring the angles of both sides in the same quadrant, adjustment by $\pi$ is to be made.

Examples

1. Evaluate $\tan ^{-1} \tan (-6)$

2. If $2 \tan ^{-1} x+\sin ^{-1} \frac{2 x}{1+x^{2}}$ is independent of $x$, then

(a). $\mathrm{x} \in[1, \infty)$

(b). $\mathrm{x} \in[-1,1]$

(c). $\mathrm{x} \in(-\infty,-1]$

(d). None of these

3. If $0<x<1$, then $\sqrt{1+x^{2}}\left[\left\{x \cos \left(\cot ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right\}^{2}-1\right]^{1 / 2}=$

(a). $\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}$

(b). $\mathrm{x}$

(c). $\mathrm{x} \sqrt{1+\mathrm{x}^{2}}$

(d). $\sqrt{1+\mathrm{x}^{2}}$

4. If $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, then the value of $\tan ^{-1}\left(\frac{\tan x}{4}\right)+\tan ^{-1}\left(\frac{3 \sin 2 x}{5+3 \cos 2 x}\right)$ is

(a). $\frac{x}{2}$

(b). $2 \mathrm{x}$

(c). $3 \mathrm{x}$

(d). $x$

5. An integral solution of the equation

$\tan ^{-1} x+\tan ^{-1} \frac{1}{y}=\tan ^{-1} 3$ is

(a). $(2,7)$

(b). $(4,-13)$

(c). $(5,-8)$

(d). $(1,2)$

6. Sum to the $\mathrm{n}$ terms of the series

$\operatorname{cosec}^{-1} \sqrt{10}+\operatorname{cosec}^{-1} \sqrt{50}+\operatorname{cosec}^{-1} \sqrt{170}+$. $+\operatorname{cosec}^{-1} \sqrt{\left(n^{2}+1\right)\left(n^{2}+2 n+2\right)}$ is

(a). $0$

(b). $\infty$

(c). $\tan ^{-1}(\mathrm{n}+1)-\frac{\pi}{4}$

(d). $\cot ^{-1}(n+1)-\frac{\pi}{4}$

7. In a $\triangle \mathrm{ABC}$, if $\mathrm{A}=\tan ^{-1} 2, \mathrm{~B}=\tan ^{-1} 3$ then $\mathrm{C}$ is equal to

(a). $\frac{\pi}{3}$

(b). $\frac{\pi}{4}$

(c). $\frac{\pi}{6}$

(d). None of these

Practice questions

1. If $x$ satisfies the inequation $x^{2}-x-2>0$, then a value exits for

(a). $\sin ^{-1} x$

(b). $\sec ^{-1} \mathrm{x}$

(c). $\cos ^{-1} \mathrm{x}$

(d). None of these

2. If $\left[\sin ^{-1} x\right]+\left[\cos ^{-1} x\right]=0$, where $x$ is a non negative real number and [.] denotes the greatest integer function, then complete set of values of $x$ is

(a). $(\cos 1,1)$

(b). $(-1, \cos 1)$

(c). $(\sin 1,1)$

(d). $(\cos 1, \sin 1)$

3. If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$, then $x y+y z+z x$ is

(a). $-3$

(b). $0$

(c). $3$

(d). $\pi$

4. If $\sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}$ then $\cos ^{-1} x+\cos ^{-1} y=$

(a). $\frac{2 \pi}{3}$

(b). $\frac{\pi}{3}$

(c). $\frac{\pi}{6}$

(d). $\pi$

5. The greatest of $\tan 1, \tan ^{-1} 1, \sin ^{-1} 1, \sin 1, \cos 1$ is

(a). $\sin 1$

(b). $\tan 1$

(c). $\tan ^{-1} 1$

(d). None of these

6. The value of $\cos ^{-1}(\cos 12)-\sin ^{-1}(\sin 12)$ is

(a). $0$

(b). $\pi$

(c). $8 \pi-24$

(d). None of these

7. $\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x$, then $\sin x=$

(a). $\tan ^{2} \frac{\alpha}{2}$

(b). $\cot ^{2} \frac{\alpha}{2}$

(c). $\tan \alpha$

(d). $\cot \frac{\alpha}{2}$

8. If $\sin ^{-1} x=2 \sin ^{-1}$ a has a solution for

(a). $|\mathrm{a}| \geq \frac{1}{\sqrt{2}}$

(b). $|\mathrm{a}| \leq \frac{1}{\sqrt{2}}$

(c). all real values of a

(d). $|\mathrm{a}|<\frac{1}{2}$

9. If $\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha$, then $4 x^{2}-4 x y \cos \alpha+y^{2}$ is equal to

(a). $2 \sin 2 \alpha$

(b). $4$

(c). $4 \sin ^{2} \alpha$

(d). $-4 \sin ^{2} \alpha$

10. If $\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\left(\frac{\pi}{2}\right)$, then the value of $x$ is

(a). 4

(b). 5

(c). 1

(d). 3

11. The value of $\cot \left(\operatorname{cosec}^{-1} \frac{5}{3}+\tan ^{-1} \frac{2}{3}\right)$ is

(a). $\frac{6}{17}$

(b). $\frac{3}{17}$

(c). $\frac{4}{17}$

(d). $\frac{5}{17}$

Assertion and Reasoning

12. $\operatorname{Let} f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$

Statement-1 : $ \mathrm{f}^{\prime}(2)=-\frac{2}{5}$ and

Statement-2:$\sin ^{-1}\left(\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\right)=\pi-2 \tan ^{-1} \mathrm{x}, \forall \mathrm{x}>1$.

(a). Statement-1 is True, statement-2 is True ;statement-2 is a correct explanation for statement-1

(b). Statement-1 is True, statement - 2 is True ; statement - 2 is NOT a correct explanation for statement - 1

(c). Statement - 1 is True, statement - 2 is False

(d). Statement - 1 is False, statement - 2 is True.

Comprehension (Que. no. 13 to 15)

$ \begin{aligned} \text { Given that } \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) & = \begin{cases}2 \tan ^{-1} x, & |x| \leq 1 \\ -\pi+2 \tan ^{-1} x, & x>1 \\ \pi+2 \tan ^{-1} x, & x<-1\end{cases} \\ \sin ^{-1}\left(\frac{2 x}{1-x^{2}}\right) & = \begin{cases}2 \tan ^{-1} x, & x \mid \leq 1 \\ \pi-2 \tan ^{-1} x, & x>1 \\ -\left(\pi+2 \tan ^{-1} x\right), & x<-1\end{cases} \end{aligned} $

and $\sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{x}=\frac{\pi}{2}$ for $-1 \leq \mathrm{x} \leq 1$.

13. $\sin ^{-1}\left(\frac{4 x}{x^{2}+4}\right)+2 \tan ^{-1}\left(-\frac{x}{2}\right)$ is independent at $x$ then :

(a). $\mathrm{x} \in[1, \infty)$

(b). $x \in[1,1]$

(c). $[-2,2]$

(d). $\mathrm{x} \in[-3,4]$

14. If $(x-1)\left(x^{2}+1\right)>0$, then $\sin \left(\frac{1}{2} \tan ^{-1} \frac{2 x}{1-x^{2}}-\tan ^{-1} x\right)=$

(a). $-1$

(b). $1$

(c). $\frac{1}{\sqrt{2}}$

(d). None of these

15. If $\cos ^{-1}\left(\frac{6 \mathrm{x}}{1+9 \mathrm{x}^{2}}\right)=-\frac{\pi}{2}+2 \tan ^{-1} 3 \mathrm{x}$ then $\mathrm{x} \in$

(a). $(-\infty,-1)$

(b). $\left(-\frac{1}{3}, \frac{1}{3}\right)$

(c). $\left(\frac{1}{3}, \infty\right)$

(d). None of these

16. Match the following