If $log_2 x ≥ 0,$ then $log_{1/\pi}\begin{Bmatrix}sin^{-1}\frac{2x}{1+x^2}+2tan^{-1}x\end{Bmatrix}$ is equal
Answer & explanation
Correct answer: option 3
We have,
$log_2 x ≥ 0 ⇒ x ≥ 2°= 1$, we have
For x ≥ 1, we have
$sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi - 2 tan^{-1}x.$
$∴ log_{1/\pi}\begin{Bmatrix}sin^{-1}\frac{2x}{1+x^2}+2tan^{-1}x\end{Bmatrix}$
$= log_{1/\pi}\begin{Bmatrix} \pi- 2 tan^{-1} x + 2 tan^{-1} x \end{Bmatrix}$
$= log_{1/\pi} \,\,\pi = - 1. $