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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Aicharya · Accepted Answer

let $t=	an 	heta$, then $\begin{aligned} & x=\cos ^{-1}\left(\frac{1}{\sqrt{1+	an ^2 	heta}}ight)=\cos ^{-1}\left(\frac{1}{\sec 	heta}ight)=\cos ^{-1} \cos 	heta=	heta \ & y=\sin ^{-1}\left(\frac{	an 	heta}{\sqrt{1+	an ^2 	heta}}ight)=\sin ^{-1}\left(\frac{	an 	heta}{\sec 	heta}ight)=\sin ^{-1} \sin 	heta=	heta \ & \Rightarrow \frac{d y}{d x}=\frac{d(	heta)}{d 	heta}=1\end{aligned}$

If $x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right), y=\sin ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$, then…

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