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Question:

If $x \sin (a+y)+\sin a \cos (a+y)=0$, prove that $\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$

Solution:

Here, $x \sin (a+y)+\sin a \cos (a+y)=0$

$x=\frac{-\sin a \cos (a+y)}{x \sin (a+y)}$

Differentiating it with respect to $x$ using the chain rule and product rule,

$\frac{d}{d x}[x \sin (a+y)+\sin a \cos (a+y)]=0$

$x \frac{d}{d x} \sin (a+y)+\sin (a+y) \frac{d x}{d x}+\sin a \frac{d}{d x} \cos (a+y)+\cos (a+y) \frac{d}{d x} \sin a=0$

$x \cos (a+y)\left(0+\frac{d y}{d x}\right)+\sin (a+y)=\sin a\left(-\sin (a+y) \frac{d y}{d x}\right)+0=0$

$[x \cos (a+y)-\sin a \sin (a+y)] \frac{d y}{d x}+\sin (a+y)=0$

$\frac{d y}{d x}=-\frac{\sin (a+y)}{x \cos (a+y)-\sin a \sin (a+y)}$

$\frac{d y}{d x}=-\frac{\sin (a+y)}{\left(\frac{-\sin a \cos (a+y)}{\sin (a+y)}\right) \cos (a+y)-\sin a \sin (a+y)}\left[\right.$ Since $\left.x=\frac{-\sin a \cos (a+y)}{x \sin (a+y)}\right]$

$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{(\sin a) \cos ^{2}(a+y)+(\sin a) \sin ^{2}(a+y)}$

$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{(\sin a)\left[\cos ^{2}(a+y)+\sin ^{2}(a+y)\right]}$

$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$

[ Since $\cos ^{2} \mathrm{a}+\sin ^{2} \mathrm{a}=1$ ]

Hence Proved.