A function $f:\mathrm{A}\to \mathrm{B}$ is invertible if $f$ it is a bijection. The inverse of $f$ is denoted by $f^{-1}$ and is defined as $f^{-1}(\mathrm{y})=\mathrm{x}\iff f(\mathrm{x})=\mathrm{y}$. 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.

Domain & range of inverse trigonometric functions

Function	Domain	Range (Principal value branch)
${\sin }^{-1}\mathrm{X}$	$[-1,1]$	$[-\pi /2,-\pi /2]$
${\cos }^{-1}\mathrm{X}$	$[-1,1]$	$[0,\pi ]$
${\tan }^{-1}\mathrm{X}$	$(-\mathrm{\infty },\mathrm{\infty })$	$(-\pi /2,\pi /2)$
${\cot }^{-1}\mathrm{X}$	$(-\mathrm{\infty },\mathrm{\infty })$	$(0,\pi )$
${\mathrm{cosec}}^{-1}\mathrm{x}$	$(-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$	$[-\pi /2,0)\cup (0,\pi /2]$
${\sec }^{-1}\mathrm{X}$	$(-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$	$[0,\pi /2)\cup (\pi /2,\pi ]$

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

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

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

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

Period $=\pi$

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

Period $=2\pi$

(vi). ${\mathrm{cosec}}^{-1}(\mathrm{cosec}x)=\{\begin{array}{l}-2n\pi +x,2n\pi -\frac{\pi }{2}\le x\le 2n\pi +\frac{\pi }{2}\\ (2n+1)\pi -x,(2n+1)\pi -\frac{\pi }{2}\le x\le (2n+1)\pi +\frac{\pi }{2},x\ne n\pi ,n\in Z\end{array}$

Period $=2\pi$

2. (i). $\sin ({\sin }^{-1}x)=x,-1\le x\le 1$ $$(ii). $\cos ({\cos }^{-1}\mathrm{x})=\mathrm{x},-1\le \mathrm{x}\le 1$

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

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

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

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

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

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

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

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

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

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

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

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

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

(v). ${\text{cosec}}^{-1}(-x)=-{\text{cosec}}^{-1}x,x\le -1$ or $x\ge 1$

(vi). ${\sec }^{-1}(-x)=\pi -{\sec }^{-1}x,x\le -1$ or $x\ge 1$

6. Conversions of inverse trigonometric functions

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

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

${\tan }^{-1}\mathrm{x}=$

$\begin{array}{rl}& {\sin }^{-1}\frac{x}{\sqrt{1+x^2}}={\cos }^{-1}\frac{1}{\sqrt{1+x^2}}={\cot }^{-1}\frac{1}{x}\\ & {\mathrm{cosec}}^{-1}\frac{\sqrt{1+x^2}}{x}={\sec }^{-1}\sqrt{1+x^2}\end{array}$

7. (i) ${\sin }^{-1}x+{\sin }^{-1}y=\{\begin{array}{c}{\sin }^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})\text{ if }-1\le x,y\le 1\mathrm{\&}{x}^2+y^2\le 1\\ \text{ or if }xy<0\mathrm{\&}{x}^2+y^2>1\\ \pi -{\sin }^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})\text{ if }0<x,y\le 1\mathrm{\&}{x}^2+y^2>1\\ -\pi -{\sin }^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})\text{ if }-1\le x,y<0\mathrm{\&}{x}^2+y^2>1\end{array}$

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

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

(ii) ${\cos }^{-1}\mathrm{x}-{\cos }^{-1}\mathrm{y}=\{\begin{array}{l}{\cos }^{-1}(\mathrm{xy}+\sqrt{1-{\mathrm{x}}^2}\sqrt{1-{\mathrm{y}}^2})\text{ if }-1\le \mathrm{x},\mathrm{y}\le 1\mathrm{\&}\mathrm{x}\le \mathrm{y}\\ -{\cos }^{-1}(\mathrm{xy}+\sqrt{1-{\mathrm{x}}^2}\sqrt{1-{\mathrm{y}}^2})\text{ if }-1\le \mathrm{y}\le 0,0<\mathrm{x}\le 1\mathrm{\&}\mathrm{x}\ge \mathrm{y}\end{array}$

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

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

Remark : If ${\mathrm{x}}_1,{\mathrm{x}}_2,\dots \dots \dots {\mathrm{x}}_{\mathrm{n}}\in \mathrm{R}$, then ${\tan }^{-1}{\mathrm{x}}_1+{\tan }^{-1}{\mathrm{x}}_2+\dots \dots \dots +{\tan }^{-1}{\mathrm{x}}_{\mathrm{n}}$ $={\tan }^{-1}\left(\frac{{\mathrm{s}}_1-{\mathrm{s}}_3+{\mathrm{s}}_5-{\mathrm{s}}_7\dots \dots \dots }{1-{\mathrm{s}}_2+{\mathrm{s}}_4-{\mathrm{s}}_6+\dots \dots .}\right)$

Where ${\mathrm{s}}_{\mathrm{k}}$ is the sum of the product of ${\mathrm{x}}_1,{\mathrm{x}}_2,\dots \dots \dots .x_n$ taken $k$ at $a$ time.

ie.

$\begin{array}{ll}{\mathrm{s}}_1=& {\mathrm{x}}_1+{\mathrm{x}}_2+\dots \dots \dots +{\mathrm{x}}_{\mathrm{n}}=\sum {\mathrm{x}}_{\mathrm{i}}\\ {\mathrm{s}}_2=& {\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{x}}_2{\mathrm{x}}_3+\dots \dots .+{\mathrm{x}}_{\mathrm{n}-1}{\mathrm{x}}_{\mathrm{n}}=\sum {\mathrm{x}}_1{\mathrm{x}}_2\\ {\mathrm{s}}_3=& \sum {\mathrm{x}}_1{\mathrm{x}}_2{\mathrm{x}}_3\dots \dots \dots \dots \dots \dots ..\mathrm{etc}.\end{array}$

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

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

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

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

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

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

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

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.

13. Hyperbolic functions

(i). $\sin h(-x)$	$=-\sin hx$	odd function
$\cos h(-\mathrm{x})$	$=-\cos h\mathrm{x}$	even function
$\tan h(-\mathrm{x})$	$=\tan h\mathrm{x}$	odd function
(ii). Function	Domain	Range
$\sin h^{-1}\mathrm{x}$	$\mathrm{R}$	$\mathrm{R}$
$\cos h^{-1}\mathrm{X}$	$(0,\mathrm{\infty })$	$(1,\mathrm{\infty })$
$\tan h^{-1}\mathrm{x}$	$\mathrm{R}$	$(-1,1)$
${\cot \mathrm{h}}^{-1}\mathrm{x}$	$\mathrm{R}-\{0\}$	$R-[-1,1]$
${\sec \mathrm{h}}^{-1}\mathrm{X}$	$(0,\mathrm{\infty })$	$(0,1)$
${\text{cosec}\mathrm{h}}^{-1}\mathrm{x}$	$\mathrm{R}-\{0\}$	$R-\{0\}$
(iii). $\sin h(\sin h^{-1}\mathrm{x})=\mathrm{x}$		$\sin h^{-1}(\sin hx)=x$
$\cos h(\cos h^{-1}\mathrm{x})=\mathrm{x}$		$\cos h^{-1}(\cos hx)=x$
$\tan h(\tan h^{-1}x)=x$		$\tan h^{-1}(\tan hx)=x$
$\sin h({\sin }^{-1}x)=x^n$		$\sin h(\sin h^{-1}{x}^n)=x^n$

(iv). $\sin h^{-1}\mathrm{x}={\log }_{\mathrm{e}}(\mathrm{x}+\sqrt{{\mathrm{x}}^2+1})$

$\cos h^{-1}x={\log }_e(x+\sqrt{x^2-1})$

$\tan h^{-1}x=\frac{1}{2}{\log }_e\left(\frac{x+1}{x-1}\right)x>1,x<-1$

${\cot \mathrm{h}}^{-1}x=\frac{1}{2}{\log }_e\left(\frac{x-1}{x+1}\right)x>1,x<-1$

${\sec \mathrm{h}}^{-1}x={\log }_e\left(\frac{1+\sqrt{1-x^2}}{x}\right)0<x\le 1$

${\text{cosec}\mathrm{h}}^{-1}x=\{\begin{array}{lll}{\log }_e\left(\frac{1+\sqrt{1+x^2}}{x}\right)& \text{ if }& x>0\\ {\log }_e\left(\frac{1-\sqrt{1+x^2}}{x}\right)& \text{ if }& x<0\end{array}$

(v). $\sin h^{-1}x={\mathrm{cosech}}^{-1}\left(\frac{1}{x}\right)$

$\sin h^{-1}x=\cos h^{-1}\sqrt{x^2+1}$

$\cos h^{-1}\mathrm{x}=\sin h^{-1}\sqrt{{\mathrm{x}}^2-1}$

$\sin h(\cos h^{-1}\mathrm{x})=\sqrt{{\mathrm{x}}^2-1}$

Solved examples

1. The sum to infinite terms of the series

${\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{13}+\dots \dots \dots \dots \dots$ is

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

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

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

(d).. None of these

Show Answer

Solution : By method of difference

$\begin{array}{rl}& \mathrm{Tn}={\tan }^{-1}\frac{1}{1+\mathrm{n}+{\mathrm{n}}^2}\\ & \mathrm{Tn}={\tan }^{-1}\frac{1}{1+n(n+1)}={\tan }^{-1}\frac{n+1-n}{n(n+1)}={\tan }^{-1}(n+1)-{\tan }^{-1}n\\ & \therefore T_n={\tan }^{-1}(n+1)-{\tan }^{-1}n\\ & {\mathrm{T}}_1^{\mathrm{n}}={\tan }^{-1}2-{\tan }^{-1}1\\ & {\mathrm{T}}_2={\tan }^{-1}3-{\tan }^{-1}2\\ & {\mathrm{T}}_3={\tan }^{-1}4-{\tan }^{-1}3\\ & .\\ & .\\ & T_n={\tan }^{-1}(n+1)-{\tan }^{-1}n\\ & \text{ Adding, }{\mathrm{S}}_{\mathrm{n}}={\mathrm{T}}_1+{\mathrm{T}}_2+\dots \dots \dots \dots +{\mathrm{T}}_{\mathrm{n}}\\ & ={\tan }^{-1}(\mathrm{n}+1)-{\tan }^{-1}1\\ & \therefore {\mathrm{S}}_{\mathrm{\infty }}={\tan }^{-1}\mathrm{\infty }-{\tan }^{-1}1=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}\end{array}$

Answer: b

2. The sum to infinite terms of the series ${\tan }^{-1}\frac{1}{{2.1}^2}+{\tan }^{-1}\frac{1}{{2.2}^2}+{\tan }^{-1}\frac{1}{{2.3}^2}+\dots \dots \dots \dots \dots \dots \mathrm{\infty }$ is

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

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

(c). $\frac{\pi }{2}$

(d). None

Show Answer

Solution :

$\begin{array}{rl}& \mathrm{Tn}={\tan }^{-1}\frac{1}{2{\mathrm{n}}^2}={\tan }^{-1}\frac{2}{4{\mathrm{n}}^2}={\tan }^{-1}\frac{(2\mathrm{n}+1)-(2\mathrm{n}-1)}{1+(2\mathrm{n}+1)(2\mathrm{n}-1)}\\ & \Rightarrow {\mathrm{T}}_{\mathrm{n}}={\tan }^{-1}(2\mathrm{n}+1)-{\tan }^{-1}(2\mathrm{n}-1)\\ & \therefore {\mathrm{T}}_1={\tan }^{-1}3-{\tan }^{-1}1\\ & {\mathrm{T}}_2={\tan }^{-1}5-{\tan }^{-1}3\\ & {\mathrm{T}}_3={\tan }^{-1}7-{\tan }^{-1}3\end{array}$

$.$

$.$

$.$

$\mathrm{Tn}={\tan }^{-1}(2n+1)-{\tan }^{-1}(2n-1)$

Adding, ${\mathrm{S}}_{\mathrm{n}}={\mathrm{T}}_1+{\mathrm{T}}_2+\dots \dots .{\mathrm{T}}_{\mathrm{n}}={\mathrm{Tan}}^{-1}(2\mathrm{n}+1)-{\tan }^{-1}1$

$\therefore {\mathrm{S}}_{\mathrm{\infty }}={\tan }^{-1}\mathrm{\infty }-{\tan }^{-1}1=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$

Answer: a

3. The value of

${\tan }^{-1}\frac{{\mathrm{c}}_1\mathrm{x}-\mathrm{y}}{{\mathrm{c}}_1\mathrm{y}+\mathrm{x}}+{\tan }^{-1}\frac{{\mathrm{c}}_2-{\mathrm{c}}_1}{1+{\mathrm{c}}_1{\mathrm{c}}_2}+{\tan }^{-1}\frac{{\mathrm{c}}_3-{\mathrm{c}}_2}{1+{\mathrm{c}}_2{\mathrm{c}}_3}+\dots \dots \dots \dots \dots \dots \dots +{\tan }^{-1}\frac{1}{{\mathrm{c}}_{\mathrm{n}}}\text{ is }$

(a). ${\tan }^{-1}\frac{x}{y}$

(b). ${\tan }^{-1}\frac{y}{x}$

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

(d). None

Show Answer

Solution: Write the series as

$\begin{array}{c}{\tan }^{-1}\frac{\frac{x}{y}-\frac{1}{c_1}}{1+\frac{x}{y}\frac{1}{c_1}}+{\tan }^{-1}\frac{\frac{1}{c_1}-\frac{1}{c_2}}{1+\frac{1}{c_1}\frac{1}{c_2}}+{\tan }^{-1}\frac{\frac{1}{c_2}-\frac{1}{c_3}}{1+\frac{1}{c_2}\frac{1}{c_3}}+\dots \dots ..+{\tan }^{-1}\frac{\frac{1}{c_{n-1}}-\frac{1}{c_n}}{1+\frac{1}{c_{n-1}}\frac{1}{c_n}}+{\tan }^{-1}\frac{1}{c_n}\\ \Rightarrow ({\tan }^{-1}\frac{x}{y}-{\tan }^{-1}\frac{1}{c_1})+({\tan }^{-1}\frac{1}{c_1}-{\tan }^{-1}\frac{1}{c_2})+({\tan }^{-1}\frac{1}{c_2}-{\tan }^{-1}\frac{1}{c_3})+\\ \dots \dots \dots \dots \dots ..+({\tan }^{-1}\frac{1}{c_{n-1}}-{\tan }^{-1}\frac{1}{c_n})+{\tan }^{-1}\frac{1}{c_n}={\tan }^{-1}\frac{x}{y}\end{array}$

Answer: a

4. The number of positive integral solutions of the equation ${\tan }^{-1}x+{\cos }^{-1}\frac{y}{\sqrt{1+y^2}}$ $={\sin }^{-1}\frac{3}{\sqrt{10}}$ is

(a). 1

(b). 2

(c). 3

(d). None

Show Answer

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

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

$\Rightarrow \frac{1}{\mathrm{y}}=\frac{3-\mathrm{x}}{1+3\mathrm{x}}$

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

Put $\mathrm{x}=1$, then $\mathrm{y}=2$ Put $\mathrm{x}=2$, then $\mathrm{y}=7$

$\therefore (1,2)\mathrm{\&}(2,7)$ are two sets.

Answer: b

5. If ${\cot }^{-1}\left(\frac{\mathrm{n}}{\pi }\right)>\frac{\pi }{6};\mathrm{n}\in \mathrm{N}$, then the maximum value of $\mathrm{n}$ can be

(a). 4

(b). 5

(c). 6

(d). None

Show Answer

Solution : $\frac{\mathrm{n}}{\pi }<\cot \frac{\pi }{6}(\because {\cot }^{-1}\mathrm{x}$ is a decreasing function $)$

$\Rightarrow \mathrm{n}<\pi \sqrt{3}$

$\mathrm{n}<5.43\Rightarrow \mathrm{n}=5(max)$

Answer: b

6. The value of

${\sin }^{-1}\{\cot ({\sin }^{-1}\left(\sqrt{\frac{2-\sqrt{3}}{4}}\right)+{\cos }^{-1}\frac{\sqrt{12}}{4}+{\sec }^{-1}\sqrt{2})\}$

(a). $0$

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

(c). $\frac{\pi }{2}$

(d). None

Show Answer

Solution : $\because {\sin }^{-1}\sqrt{\frac{2-\sqrt{3}}{4}}={\sin }^{-1}\sqrt{\frac{4-2\sqrt{3}}{8}}$

$\begin{array}{rl}& ={\sin }^{-1}\sqrt{\frac{3+1-2\sqrt{3}}{(2\sqrt{2}{)}^2}}\\ & ={\sin }^{-1}\sqrt{{\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)}^2}\\ & ={\sin }^{-1}\frac{\sqrt{3}-1}{2\sqrt{2}}\end{array}$

$\begin{array}{rl}& ={\sin }^{-1}\sin \frac{\pi }{12}=\frac{\pi }{12}\\ \therefore {\sin }^{-1}\{\cot (\frac{\pi }{12}+\frac{\pi }{6}+\frac{\pi }{4})\}& ={\sin }^{-1}\cot \frac{\pi }{2}\\ & ={\sin }^{-1}0=0\end{array}$

Answer: a

7. The greatest value of ${({\tan }^{-1}x)}^2+{({\cot }^{-1}x)}^2$ is…………..

Show Answer

Solution :

${({\tan }^{-1}x)}^2+{({\cot }^{-1}x)}^2={({\tan }^{-1}x+{\cot }^{-1}x)}^2-2{\tan }^{-1}x{\cot }^{-1}x$

$=\frac{{\pi }^2}{4}-2{\tan }^{-1}x(\frac{\pi }{2}-{\tan }^{-1}\mathrm{x})$

Let ${\tan }^{-1}\mathrm{x}=\mathrm{y}$, then LHS $=\frac{{\pi }^2}{4}-2\mathrm{y}(\frac{\pi }{2}-\mathrm{y})$

$\begin{array}{rl}& =\frac{{\pi }^2}{4}-\pi y+2y^2=2(y^2-\frac{\pi y}{2}+\frac{{\pi }^2}{16})-\frac{2\cdot {\pi }^2}{16}+\frac{{\pi }^2}{4}\\ & =2{({\tan }^{-1}x-\frac{\pi }{4})}^2+\frac{{\pi }^2}{8}\end{array}$

$\therefore$ Minimum value is $\frac{{\pi }^2}{8}$

Answer : $\frac{{\pi }^2}{8}$

Practice questions

1. If ${({\tan }^{-1}x)}^2+{({\cot }^{-1}x)}^2=\frac{5{\pi }^2}{8}$, then $x$ equals

(a). $-1$

(b). $1$

(c). $0$

(d). None of these

Show Answer

Answer: (a).

2. If ${\sin }^{-1}(x-\frac{x^2}{2}+\frac{x^3}{4}-\dots \dots ..)+{\cos }^{-1}(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\dots \dots \dots .)=\frac{\pi }{2}$ for $0<|x|<\sqrt{2}$, then $x$ equals

(a). $\frac{1}{2}$

(b). $1$

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

(d). $-1$

Show Answer

Answer: (b).

3. Match the conditions / expressions in column I with statement in column II.

Let $(\mathrm{x},\mathrm{y})$ be such that ${\sin }^{-1}(\mathrm{ax})+{\cos }^{-1}\mathrm{y}+{\cos }^{-1}(\mathrm{bxy})=\frac{\pi }{2}$

Column I	Column II
(a). If $a=1\mathrm{\&}b=0$, then( $x,y)$	(p). lies on the circle $x^2+y^2=1$
(b). If $a=1\mathrm{\&}b=1$, then $(x,y)$	(q). lies on $(x^2-1)(y^2-1)=0$
(c). If $a=1\mathrm{\&}b=2$, then $(x,y)$	(r). lies on $y=x$
(b). If $a=2\mathrm{\&}b=2$, then( $x,y)$	(s). lies on $(4x^2-1)(y^2-1)=0$

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Answer: a $\text{rarr}$ p, b $\text{rarr}$ q, c $\text{rarr}$ p, d $\text{rarr}$ s

4. Sum to $n$ terms of the series

${\mathrm{cosec}}^{-1}\sqrt{10}+{\mathrm{cosec}}^{-1}\sqrt{50}+{\mathrm{cosec}}^{-1}\sqrt{170}+\dots \dots \dots +{\mathrm{cosec}}^{-1}\sqrt{(n^2+1)(n^2+2n+2)}\text{ is }$

(a). $0$

(b). $\mathrm{\infty }$

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

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

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Answer: (c).

5. Match the following.

Let $t_1={({\sin }^{-1}x)}^{{\sin }^{-1}x},t_2={({\sin }^{-1}x)}^{{\cos }^{-1}x},t_3={({\cos }^{-1}x)}^{{\sin }^{-1}x},t_4={({\cos }^{-1}x)}^{{\cos }^{-1}x}$

Column I	Column II
(a). $\mathrm{x}\in (0,\cos 1)$	(p). ${\mathrm{t}}_1>{\mathrm{t}}_2>{\mathrm{t}}_4>{\mathrm{t}}_3$
(b). $\mathrm{x}\in (\cos 1,\frac{1}{\sqrt{2}})$	(q). ${\mathrm{t}}_4>{\mathrm{t}}_3>{\mathrm{t}}_1>{\mathrm{t}}_2$
(c). $\mathrm{x}\in (\frac{1}{\sqrt{2}},\sin 1)$	(r). ${\mathrm{t}}_1>{\mathrm{t}}_2>{\mathrm{t}}_4>{\mathrm{t}}_3$
(c). $\mathrm{x}\in (\sin 1,1)$	(s). $t_3>t_4>t_1>t_2$

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Answer: a $\text{rarr}$ q, b $\text{rarr}$ s, c $\text{rarr}$ r, d $\text{rarr}$ s

6. Read the passage & answer the following questions

If ${\tan }^{-1}x:{\tan }^{-1}\mathrm{y}=1:4($ where $|\mathrm{x}|<\tan \frac{\pi }{6}$ ) then

(i). The value of $y$ as an algebraic function of $x$ will be

(a). $\frac{4x(1+x^2)}{x^4-6x^2+1}$

(b). $\frac{4x(1-x^2)}{x^4-6x^2+1}$

(c). $\frac{4x(1+x^2)}{x^4+6x^2+1}$

(d). None of these.

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Answer: (b).

(ii). The root of the equation $x^4-6x^2+1=0$ is

(a). $\tan \frac{\pi }{12}$

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

(c). $\tan \frac{\pi }{8}$

(d). $\tan \frac{\pi }{16}$

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Answer: (a).

7. If a ${\sin }^{-1}\mathrm{x}-\mathrm{b}{\cos }^{-1}\mathrm{x}=\mathrm{c}$, then $a{\sin }^{-1}\mathrm{x}+\mathrm{b}{\cos }^{-1}\mathrm{x}$ is

(a). $0$

(b). $\frac{\pi \mathrm{ab}+\mathrm{c}(\mathrm{b}-\mathrm{a})}{\mathrm{a}+\mathrm{b}}$

(c). $\frac{\pi }{2}$

(d). $\frac{\pi \mathrm{ab}+\mathrm{c}(\mathrm{b}-\mathrm{a})}{\mathrm{a}+\mathrm{b}}$

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Answer: (d).

8. $\sum _{\mathrm{r}=1}^{\mathrm{n}}{\mathrm{Sin}}^{-1}\left(\frac{\sqrt{\mathrm{r}}-\sqrt{\mathrm{r}-1}}{\sqrt{\mathrm{r}(\mathrm{r}+1)}}\right)$ is

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

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

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

(d). ${\tan }^{-1}\sqrt{\mathrm{n}+1}$

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Answer: (c).

9. If $[{\cot }^{-1}x]+[{\cos }^{-1}x]=0$, then complete set of values of $x$ is

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

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

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

(d). None of these

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Answer: (c).

10. If $({\sin }^{-1}x+{\sin }^{-1}w)({\sin }^{-1}y+{\sin }^{-1}z)={\pi }^2$, then

$D=\left|\begin{array}{ll}{x}^{N}1& y^{N}2\\ z^N& w^{N}4\end{array}\right|$ where $N_1,N_2,N_3,N_4\in W$

(a). has a maximum value of $2$

(b). has a minimum value of $0$

(c). 16 different $\mathrm{D}$ are possible

(d). has a minimum value of $-2.$

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Answer: (a, c, d)

11. The value of $k(k>0)$ such that the length of the longest interval in which the function $f(x)={\sin }^{-1}|\sin kx|+{\cos }^{-1}(\cos kx)$ is constant is $\frac{\pi }{4}$ is $/$ are

(a). 8

(b). 4

(c). 12

(d). 16

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Answer: (b).

12. Match the following

Column I	Column II
(a). ${({\sin }^{-1}\mathrm{x})}^2+{({\sin }^{-1}\mathrm{y})}^2=\frac{{\pi }^2}{2\Rightarrow {\mathrm{x}}^3+{\mathrm{y}}^3}=$	(p). $1$
(b). ${({\cos }^{-1}\mathrm{x})}^2+{({\cos }^{-1}\mathrm{y})}^2=2{\pi }^{2\Rightarrow {\mathrm{x}}^5+{\mathrm{y}}^5}=$	(q). $-2$
(c). ${({\sin }^{-1}\mathrm{x})}^2+{({\sin }^{-1}\mathrm{y})}^2=\frac{{\pi }^4}{4}\Rightarrow \mid \mathrm{x}-\mathrm{y}\mid =$	(r). $0$
(d). $\mid {\sin }^{-1}x-{\sin }^{-1}y\mid =\pi \Rightarrow {\mathrm{x}}^{\mathrm{y}}=$	(s). $2$

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Answer: a $\text{rarr}$ q, r, s; b $\text{rarr}$ q; c $\text{rarr}$ r, s; d $\text{rarr}$ p