TRIGONOMETRY FUNCTIONS - 3 (Inverse Trigonometric Functions - Problem Solving)

A function $f: \mathrm{A} \rightarrow \mathrm{B}$ 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.

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

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

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). $\tan \left(\tan ^{-1} \mathrm{x}\right)=\mathrm{x}, \mathrm{x} \in \mathrm{R}$ $\hspace{1cm}$(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). $\operatorname{\cosec}\left(\operatorname{\cosec}^{-1} \mathrm{x}\right)=\mathrm{x}, \mathrm{x} \in \mathrm{R}(-\infty,-1] \cup[1, \infty)$

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

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

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

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

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

(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} \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}}=\operatorname{cosec}^{-1} \frac{1}{\mathrm{x}}=\sec ^{-1} \frac{1}{\sqrt{1-\mathrm{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} $

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

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

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

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

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

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

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.

$ \begin{array}{ll} \mathrm{s} _{1}= & \mathrm{x} _{1}+\mathrm{x} _{2}+\ldots \ldots \ldots+\mathrm{x} _{\mathrm{n}}=\sum \mathrm{x} _{\mathrm{i}} \\ \mathrm{s} _{2}= & \mathrm{x} _{1} \mathrm{x} _{2}+\mathrm{x} _{2} \mathrm{x} _{3}+\ldots \ldots .+\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} \ldots \ldots \ldots \ldots \ldots \ldots . . \mathrm{etc.} \end{array} $

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

(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.

13. Hyperbolic functions

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

$\cos h ^{-1} x=\log _{e}\left(x+\sqrt{x^{2}-1}\right)$

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

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

$\operatorname{\sec h}^{-1} x=\log _{e}\left(\frac{1+\sqrt{1-x^{2}}}{x}\right) 0<x \leq 1$

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

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

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

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

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

Solved examples

1. The sum to infinite terms of the series

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

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

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

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

(d).. None of these

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

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

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

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

(d). None

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}}+\ldots \ldots \ldots \ldots \ldots \ldots \ldots+\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

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

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

6. The value of

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

(a). $0$

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

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

(d). None

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

Practice questions

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

(a). $-1 $

(b). $1$

(c). $0$

(d). None of these

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

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

(b). $1$

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

(d). $-1$

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

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

4. Sum to $n$ terms of the series

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

(a). $0$

(b). $\infty$

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

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

5. Match the following.

Let $t _{1}=\left(\sin ^{-1} x\right)^{\operatorname{\sin}^{-1} x}, t _{2}=\left(\sin ^{-1} x\right)^{\cos ^{-1} x}, t _{3}=\left(\cos ^{-1} x\right)^{\sin ^{-1} x}, t _{4}=\left(\cos ^{-1} x\right)^{\cos ^{-1} x}$

6. Read the passage & answer the following questions

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

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

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

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

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

(d). None of these.

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

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

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

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

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

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

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

9. If $\left[\cot ^{-1} x\right]+\left[\cos ^{-1} x\right]=0$, 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

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

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

(a). 8

(b). 4

(c). 12

(d). 16

12. Match the following