Question

If $f\left(x\right)={\tan }^{-1}(\text{cosec}\left({\tan }^{-1}x\right)-\tan \left({\cot }^{-1}x\right));x>0,$ then the value of $8f^{\prime }\left(1\right)$ is

Answer 1

Given :
$f\left(x\right)={\tan }^{-1}(\text{cosec}\left({\tan }^{-1}x\right)-\tan \left({\cot }^{-1}x\right))$
$={\tan }^{-1}[\text{cosec}\left({\text{cosec}}^{-1}\frac{\sqrt{1+x^2}}{x}\right)-\tan \left({\tan }^{-1}\frac{1}{x}\right)]$
$={\tan }^{-1}[\frac{\sqrt{1+x^2}}{x}-\frac{1}{x}]$
$={\tan }^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}\right]$
Putting $x=\tan \theta$
$={\tan }^{-1}\left[\frac{\sec \theta -1}{\tan \theta }\right]$
$={\tan }^{-1}\left[\frac{1-\cos \theta }{\sin \theta }\right]$
$={\tan }^{-1}\left[\frac{2{\sin }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cdot \cos \frac{\theta }{2}}\right]$
$={\tan }^{-1}\tan \frac{\theta }{2}=\frac{\theta }{2}$
$\therefore f\left(x\right)=\frac{1}{2}{\tan }^{-1}x$
Differentiating both sides w.r.t. $x$
$f^{\prime }\left(x\right)=\frac{1}{2(1+x^2)}$
$\Rightarrow f^{\prime }\left(1\right)=\frac{1}{4}$
$\therefore 8f^{\prime }\left(1\right)=2$