If $y=\left(\log _{\cos x} \sin x\right)\left(\log _{\sin x} \cos x\right)+\sin ^{-1} \frac{2 x}{1+x^2}$, then $\frac{d y}{d x}$ at $x=\frac{\pi}{2}$ is equal to
Answer & explanation
Correct answer: option 3
We have,
$y=\left(\log _{\cos x} \sin x\right) \times\left(\log _{\sin x} \cos x\right)+\sin ^{-1} \frac{2 x}{1+x^2}$
$\Rightarrow y= \begin{cases}1-\pi-2 \tan ^{-1} x, & x<-1 \\ 1+2 \tan ^{-1} x, & \text { if }-1 \leq x \leq 1 \\ 1+\pi-2 \tan ^{-1} x, & \text { if } x>1\end{cases}$
$\Rightarrow \frac{d y}{d x}=\left\{\begin{array}{l}\frac{-2}{1+x^2}, x<-1 \text { or } x>1 \\ \frac{2}{1+x^2},-1<x<1\end{array}\right.$
$\Rightarrow \left(\frac{d y}{d x}\right)_{x=\pi / 2}=\frac{-2}{1+\frac{\pi^2}{4}}=\frac{-8}{4+\pi^2}$