Question

If $f\left(x\right)=\left({\log }_{\cot x}\tan x\right)\left({\log }_{\tan x}\cot x\right)+{\tan }^{-1}\frac{4x}{4-x^2}$ then $2f^{\prime }\left(2\right)$ is equal to

Answer 1

$f\left(x\right)=\left({\log }_{\cot x}\tan x\right)\left({\log }_{\tan x}\cot x\right)+{\tan }^{-1}\frac{4x}{4-x^2}$

$=\frac{\log \cot x}{\log \tan x}.\frac{\log \tan x}{\log \cot x}+{\tan }^{-1}\frac{4x}{4-x^2}=1++{\tan }^{-1}\frac{4x}{4-x^2}$

put $x=2\cot \theta$

$f\left(x\right)=1+{\tan }^{-1}\frac{4\cdot 2\cot \theta }{4(1-{\cot }^2\theta )}=1+{\tan }^{-1}\frac{-2\sin \theta \cos \theta }{{\cos }^2\theta -{\sin }^2\theta }=1+{\tan }^{-1}\tan (-2\theta )$

$\Rightarrow f\left(x\right)=1-2\theta =1-2{\cot }^{-1}\left(\frac{x}{2}\right)$

$\Rightarrow f^{\prime }\left(x\right)=2\frac{1}{1+{\left(\frac{x}{2}\right)}^2}\frac{1}{2}=\frac{1}{1+{\left(\frac{x}{2}\right)}^2}$

$\Rightarrow 2f^{\prime }\left(2\right)=1$