Answer
(a)$sin^{-1}x=tan^{-1} \frac{x}{\sqrt {1-x^2}}$, $\vert{x}\vert<1$
(b)$cos^{-1}x=\frac{\pi}{2}-tan^{-1}\frac{x}{\sqrt {1-x^2}}$,$\vert{x}\vert<1$
Work Step by Step
(a)Let
$tan^{-1} \frac{x}{\sqrt {1-x^2}}=\theta$ ........................... eq (1)
where $\vert{x}\vert<1$
$ \frac{x}{\sqrt {1-x^2}}=tan\theta$
Squaring
$ (\frac{x}{\sqrt {1-x^2}})^2=(tan\theta)^2$
$\frac{x^2}{( \sqrt {1-x^2})^2}=tan^2\theta$
$\frac{x^2}{1-x^2}=tan^2\theta$
$\frac{x^2}{1-x^2}+1=1+tan^2\theta$
$\frac{1}{1-x^2}=sec^2\theta$
$\frac{1}{1-x^2}=\frac{1}{cos^2\theta}$
$cos^2\theta=1-x^2$
$x^2=1-cos^2\theta$
$x^2=sin^2\theta$
Taking positive square root
$x=sin\theta$
$sin\theta=x$
$\theta=sin^{-1}x$ ..................... eq (2)
From equation (1) and equation (2)
$sin^{-1}x=tan^{-1} \frac{x}{\sqrt {1-x^2}}$, $\vert{x}\vert<1$
(b)
Let
$sin^{-1}x=\alpha$ .......................... eq (3)
$x=sin\alpha$
$x=cos(\frac{\pi}{2}-\alpha)$
$cos^{-1}x=\frac{\pi}{2}-\alpha$ ........................ eq (4)
From equation (3) and equation (4)
$cos^{-1}x=\frac{\pi}{2}- sin^{-1}x $ ........................ eq (5)
From the the equation proved in the result (a)
$sin^{-1}x=tan^{-1} \frac{x}{\sqrt {1-x^2}}$, $\vert{x}\vert<1$ ..................... eq (6)
From equation (5) and equation (6)
$cos^{-1}x=\frac{\pi}{2}-tan^{-1}\frac{x}{\sqrt {1-x^2}}$,$\vert{x}\vert<1$
