If tan x = frac{2t}{1-t^2} and sin y = frac{2 | vedclass

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(A) Given,$	an x = \frac{2t}{1-t^2}$ and $\sin y = \frac{2t}{1+t^2}$.We know the trigonometric identities: $	an(2	heta) = \frac{2	an	heta}{1-	an

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(A) Given,$	an x = \frac{2t}{1-t^2}$ and $\sin y = \frac{2t}{1+t^2}$.We know the trigonometric identities: $	an(2	heta) = \frac{2	an	heta}{1-	an^2	heta}$ and $\sin(2	heta) = \frac{2	an	heta}{1+	an^2	heta}$.Let $t = 	an	heta$,then:$x = 	an^{-1}\left(\frac{2	an	heta}{1-	an^2	heta}ight) = 	an^{-1}(	an 2	heta) = 2	heta = 2	an^{-1}t$.$y = \sin^{-1}\left(\frac{2	an	heta}{1+	an^2	heta}ight) = \sin^{-1}(\sin 2	heta) = 2	heta = 2	an^{-1}t$.Thus,$x = 2	an^{-1}t$ and $y = 2	an^{-1}t$.This implies $x = y$,so $\frac{dy}{dx} = 1$.

If $\tan x = \frac{2t}{1-t^2}$ and $\sin y = \frac{2t}{1+t^2}$,then the value of $\frac{dy}{dx}$ is

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