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

Question

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

Aicharya · Accepted Answer

Given $\sin \left(\frac{x+y}{x-y}ight)=	an \frac{\pi}{5}$ $	herefore \frac{x+y}{x-y}=\sin ^{-1}\left(	an \frac{\pi}{5}ight) \Rightarrow \frac{x+y}{x-y}=K \quad \ldots$ say ...(1) $	herefore x+y=K(x-y)$ $\quad 1+\frac{d y}{d x}=K\left(1-\frac{d y}{d x}ight) \Rightarrow 1+\frac{d y}{d x}=K-K \frac{d y}{d x}$ $(1+K) \frac{d y}{d x}=K-1 \Rightarrow \frac{d y}{d x}=\frac{K-1}{K+1}$ $\begin{aligned} \frac{d y}{d x} &=\frac{\frac{x+y}{x-y}-1}{\frac{x+y}{x-y}+1} \quad[\ldots 	ext { From (1) }] \ &=\frac{x+y-x+y}{x+y+x-y}=\frac{2 y}{2 x}=\frac{y}{x} \end{aligned}$ This problem can also be solved as follow: $\frac{x+y}{x-y}=\sin ^{-1}\left(	an \frac{\pi}{5}ight)$ Differentiating w.r.t. $\mathrm{x}$, we get $\begin{aligned} & \frac{(x-y)\left(1+\frac{d y}{d x}ight)-(x+y)\left(1-\frac{d y}{d x}ight)}{(x-y)^{2}}=0 \ 	herefore &\left(x-y+x \frac{d y}{d x}-y \frac{d y}{d x}ight)-\left(x+y-x \frac{d y}{d x}-y \frac{d y}{d x}ight)=0 \ 	herefore &-2 y+2 x \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{y}{x} \end{aligned}$

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

Solution