Question

If $f\left(x\right)=\sin [{\tan }^{-1}\left(\frac{1-x^2}{2x}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)]\forall x\in (0,1),$ then the value of $f^{\prime }\left(x\right)$ is

Answer 1

Given :
$f\left(x\right)=\sin [{\tan }^{-1}\left(\frac{1-x^2}{2x}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)]$
Putting $x=\tan \theta$
$=\sin [{\tan }^{-1}\left(\frac{1-{\tan }^2\theta }{2\tan \theta }\right)+{\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)]$
$=\sin \left[{\tan }^{-1}\right(\cot 2\theta )+{\cos }^{-1}(\cos 2\theta \left)\right]$
$=\sin [{\tan }^{-1}\tan (\frac{\pi }{2}-2\theta )+{\cos }^{-1}(\cos 2\theta \left)\right]$

As $2\theta \in (0,\frac{\pi }{2})$, so the above expression can be written as
$f\left(x\right)=\sin (\frac{\pi }{2}-2\theta +2\theta )$
$f\left(x\right)=\sin \frac{\pi }{2}=1$
$\therefore f^{\prime }\left(x\right)=0$