Answer:
Option A
Explanation:
Let y = $\sin^{-1}\left(\frac{t}{\sqrt{1+t^{2}}}\right)$
put $t= \tan\theta\Rightarrow\theta=\tan^{-1}t$
= $\sin^{-1}\left(\frac{\tan\theta}{\sqrt{1+\tan^{2}\theta}}\right)=\sin^{-1}\left(\frac{\tan\theta}{\sec\theta}\right)$
=$\sin^{-1}(\sin\theta)=\theta=\tan^{-1}t$
and $z=\cos^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)=\cos^{-1}\left(\frac{1}{\sqrt{1+\tan^{2}\theta}}\right)=\cos^{-1}(\cos \theta)$
$\theta=\tan^{-1}t$
$\therefore$ $\frac{dy}{dz}=\frac{\frac{dy}{dt}}{\frac{dz}{st}}$
$=\frac{\frac{1}{(1+t^{2})}}{\frac{1}{(1+t^{2})}}=1$
