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

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(C) Given,$x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}ight)$ and $y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}ight)$.Let $t = 	an 	heta$,then $	heta =

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(C) Given,$x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}ight)$ and $y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}ight)$.Let $t = 	an 	heta$,then $	heta = 	an^{-1} t$.For $x$,we have $x = \cos^{-1}\left(\frac{1}{\sqrt{1+	an^2 	heta}}ight) = \cos^{-1}\left(\frac{1}{\sec 	heta}ight) = \cos^{-1}(\cos 	heta) = 	heta = 	an^{-1} t$.For $y$,we have $y = \sin^{-1}\left(\frac{	an 	heta}{\sqrt{1+	an^2 	heta}}ight) = \sin^{-1}\left(\frac{	an 	heta}{\sec 	heta}ight) = \sin^{-1}(\sin 	heta) = 	heta = 	an^{-1} t$.Since $x = 	an^{-1} t$ and $y = 	an^{-1} t$,it follows that $y = x$.Therefore,differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx}(x) = 1$.

$x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right), y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right) \Rightarrow \frac{d y}{d x}$ is equal to

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