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If $x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right), y=\sin ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ is
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let $t=\tan \theta$, then
$\begin{aligned} & x=\cos ^{-1}\left(\frac{1}{\sqrt{1+\tan ^2 \theta}}\right)=\cos ^{-1}\left(\frac{1}{\sec \theta}\right)=\cos ^{-1} \cos \theta=\theta \\ & y=\sin ^{-1}\left(\frac{\tan \theta}{\sqrt{1+\tan ^2 \theta}}\right)=\sin ^{-1}\left(\frac{\tan \theta}{\sec \theta}\right)=\sin ^{-1} \sin \theta=\theta \\ & \Rightarrow \frac{d y}{d x}=\frac{d(\theta)}{d \theta}=1\end{aligned}$
$\begin{aligned} & x=\cos ^{-1}\left(\frac{1}{\sqrt{1+\tan ^2 \theta}}\right)=\cos ^{-1}\left(\frac{1}{\sec \theta}\right)=\cos ^{-1} \cos \theta=\theta \\ & y=\sin ^{-1}\left(\frac{\tan \theta}{\sqrt{1+\tan ^2 \theta}}\right)=\sin ^{-1}\left(\frac{\tan \theta}{\sec \theta}\right)=\sin ^{-1} \sin \theta=\theta \\ & \Rightarrow \frac{d y}{d x}=\frac{d(\theta)}{d \theta}=1\end{aligned}$
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