Match the following

A	The inverse function of  $\sin \left({\mathrm{Tan}}^{-1}x\right)$ is	p	$\cot \left({\cos }^{-1}x\right)$
B	The inverse function of  $f\left(x\right)=1-2^{-x}$ is	q	$\frac{{\log }_{2}x}{\left({\log }_{2}x\right)-1}$
C	The inverse function of $f\left(x\right)=2^{\frac{x}{x-1}}$	r	$\tan \left({\sin }^{-1}x\right)$
		s	$-{\log }_2(1-x)$
		t	$\frac{{\displaystyle x}}{\sqrt{1-x^2}}$

(A)As $\sin \left({\tan }^{-1}x\right)=y\Rightarrow x=\tan \left({\sin }^{-1}y\right)$

$\therefore$ $f^{-1}\left(x\right)=\tan \left({\sin }^{-1}x\right)$

$f\left(x\right)=\sin \left({\tan }^{-1}x\right)=\sin \left({\sin }^{-1}\frac{x}{\sqrt{1+x^2}}\right)=\frac{x}{\sqrt{1+x^2}}$

$y=\frac{x}{\sqrt{1+x^2}}\Rightarrow x=\frac{y}{\sqrt{1-y^2}}$

$f^{-1}\left(x\right)=\frac{{\displaystyle x}}{\sqrt{1-x^2}}$

$\tan \left({\sin }^{-1}x\right)=\mathrm{co}t\left(\frac{\pi }{2}-{\sin }^{-1}x\right)=\cot \left({\cos }^{-1}x\right)=\frac{x}{\sqrt{1-x^2}}$

(B)  $f\left(x\right)=y=1-2^{-x}$

$$\begin{gathered} 2^{-x}=1-y\Rightarrow 2^x=\frac{1}{1-y} \\ \therefore x={\log }_2\left(\frac{1}{1-y}\right) \end{gathered}$$

$f^{-1}\left(x\right)={\log }_2\frac{1}{(1-x)}$

$f^{-1}\left(x\right)=-{\log }_2(1-x)$

(C)  $f\left(x\right)=y=2^{\frac{x}{x-1}}\Rightarrow {\log }_2 y=\frac{x}{x-1}$

$x{\log }_2 y-{\log }_2 y=x$

$x\left({\log }_2 y-1\right)={\log }_2 y$

$x=\frac{{\log }_2 y}{{\log }_2 y-1}$

$f^{-1}\left(x\right)=\frac{{\log }_{2}x}{\left({\log }_{2}x\right)-1}$