If cos x = frac{1}{sqrt{1 + t^2}} and sin y = | vedclass

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(D) Given $\cos x = \frac{1}{\sqrt{1 + t^2}}$. Let $t = 	an 	heta$,then $\cos x = \frac{1}{\sqrt{1 + 	an^2 	heta}} = \frac{1}{\sec 	heta} = \cos

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(D) Given $\cos x = \frac{1}{\sqrt{1 + t^2}}$. Let $t = 	an 	heta$,then $\cos x = \frac{1}{\sqrt{1 + 	an^2 	heta}} = \frac{1}{\sec 	heta} = \cos 	heta$. Thus,$x = 	heta = 	an^{-1} t$.Given $\sin y = \frac{t}{\sqrt{1 + t^2}}$. Substituting $t = 	an 	heta$,we get $\sin y = \frac{	an 	heta}{\sec 	heta} = \sin 	heta$. Thus,$y = 	heta = 	an^{-1} t$.Since $x = 	an^{-1} t$ and $y = 	an^{-1} t$,we have $x = y$.Therefore,$\frac{dy}{dx} = \frac{d}{dx}(x) = 1$.

If $\cos x = \frac{1}{\sqrt{1 + t^2}}$ and $\sin y = \frac{t}{\sqrt{1 + t^2}}$,then $\frac{dy}{dx} = $

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