If x=cos ^{-1}left(frac{1}{sqrt{1+t^2}}right) | vedclass

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(C) Let $t = 	an 	heta$. Then $	heta = 	an^{-1} t$. $x = \cos^{-1}\left(\frac{1}{\sqrt{1+	an^2 	heta}}ight) = \cos^{-1}\left(\frac{1}{\sec 	h

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(C) Let $t = 	an 	heta$. Then $	heta = 	an^{-1} t$. $x = \cos^{-1}\left(\frac{1}{\sqrt{1+	an^2 	heta}}ight) = \cos^{-1}\left(\frac{1}{\sec 	heta}ight) = \cos^{-1}(\cos 	heta) = 	heta$. $y = \sin^{-1}\left(\frac{	an 	heta}{\sqrt{1+	an^2 	heta}}ight) = \sin^{-1}\left(\frac{	an 	heta}{\sec 	heta}ight) = \sin^{-1}(\sin 	heta) = 	heta$. Since $x = 	heta$ and $y = 	heta$,we have $y = x$. Therefore,$\frac{dy}{dx} = \frac{d}{dx}(x) = 1$.

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

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