Question

If ${\log }_{\pi }x>0$ then the absolute value of ${\log }_{\frac{1}{\pi }}\left({\sin }^{-1}\frac{2x}{1+x^2}+2{\tan }^{-1}x\right)$ is equal to

Answer 1

Answer: 1

Since ${\log }_{\pi }x>0\Rightarrow x>1$
For $x>1,{\sin }^{-1}\frac{2x}{1+x^2}=\pi -2{\tan }^{-1}x$
${\log }_{\frac{1}{\pi }}\left(\pi -2{\tan }^{-1}x+2{\tan }^{-1}x\right)$
$={\log }_{\frac{1}{\pi }}(\pi )=-1$