If sin left(frac{x+y}{x-y}right)=tan frac{pi} | vedclass

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(B) Given $\sin \left(\frac{x+y}{x-y}ight)=	an \frac{\pi}{5}$.Let $K = \sin^{-1}(	an \frac{\pi}{5})$,which is a constant.Then $\frac{x+y}{x-y} = K

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$\frac{y}{x}$

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(B) Given $\sin \left(\frac{x+y}{x-y}ight)=	an \frac{\pi}{5}$.Let $K = \sin^{-1}(	an \frac{\pi}{5})$,which is a constant.Then $\frac{x+y}{x-y} = K$.$x+y = K(x-y)$.Differentiating both sides with respect to $x$:$1 + \frac{dy}{dx} = K(1 - \frac{dy}{dx})$.$1 + \frac{dy}{dx} = K - K\frac{dy}{dx}$.$\frac{dy}{dx}(1+K) = K-1$.$\frac{dy}{dx} = \frac{K-1}{K+1}$.Substituting $K = \frac{x+y}{x-y}$ back into the expression:$\frac{dy}{dx} = \frac{\frac{x+y}{x-y} - 1}{\frac{x+y}{x-y} + 1} = \frac{x+y - (x-y)}{x+y + (x-y)} = \frac{2y}{2x} = \frac{y}{x}$.

If $\sin \left(\frac{x+y}{x-y}\right)=\tan \frac{\pi}{5}$,then $\frac{d y}{d x}=$

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