Question

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

Answer 1

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

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

i.e.,$\frac{dx}{dy}=-\sin a\times \frac{d}{dy}\left(\frac{\cos (a+y)}{\sin (a+y)}\right)$ ...[Differentiating w.r.t y both the sides]

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

$\frac{dx}{dy}=-\sin a\times \frac{-\sin (a+y)\sin (a+y)-\cos (a+y)\cos (a+y)}{\left[\sin \right(a+y){]}^2}=\sin a\times \left[\frac{{\sin }^2(a+y)+{\cos }^2(a+y)}{\left[\sin \right(a+y){]}^2}\right]$

$\frac{dx}{dy}=\sin a\times \frac{1}{{\sin }^2(a+y)}$

Therefore, $\frac{dy}{dx}=\frac{{\sin }^2(a+y)}{\sin a}$

Hence proved.