Question

$limx\to 0\frac{co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)}{si{n}^{-1}x}.$

Answer 1

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

$x\to 0$

Applying $L-$ Hospital rule

It

$=\frac{\frac{-1}{\sqrt{1-{\left(\frac{1-x^2}{1+x^2}\right)}^2}}\frac{\left[\right(1+x^2\left)\right(-2x)-(1-x^2\left)\right(2x\left)\right]}{(1+x^2{)}^2}}{\frac{1}{\sqrt{1-x^2}}}$

$x\to 0$

It

$=\sqrt{\frac{1-x^2}{4x^2}}-4x$

$x\to 0$

It

$=\frac{\sqrt{1-x^2}}{2x}.-4x$

$=-2$