Question

Solve:${\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

Answer 1

Consider the given function $={\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

Let, put $x=\tan A$ then $A={\tan }^{-1}x$ ……. (1)

Then,

${\sin }^{-1}\left(\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}\right)$

$={\sin }^{-1}\left(\frac{1-\frac{{\sin }^{2}A}{{\cos }^{2}A}}{1+\frac{{\sin }^{2}A}{{\cos }^{2}A}}\right)$

$={\sin }^{-1}\left(\frac{{\cos }^{2}A-{\sin }^{2}A}{{\cos }^{2}A+{\sin }^{2}A}\right)$

$={\sin }^{-1}\left(\frac{\cos 2A}{1}\right)$

$={\sin }^{-1}\cos 2A$

$={\sin }^{-1}\sin ({90}^0-2A)$

$={90}^0-2A$

By equation (1) and we get,

$={90}^0-2{\tan }^{-1}x$

Hence, this is the answer.