Question

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

Answer 1

$\sin y=x\sin (a+y)$

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

Differentiating with sides w.r.t.$y$, we get

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

$\Rightarrow \frac{dx}{dy}=\frac{\cos y(\sin a\cos y+cosa\sin y)-\sin y(\cos a\cos y-\sin a\sin y)}{{\sin }^2(a+y)}$

$\Rightarrow \frac{dx}{dy}=\frac{{\cos }^{2}y\sin a+cosa\sin y\cos y-\sin y\cos a\cos y+\sin a{\sin }^{2}y}{{\sin }^2(a+y)}$

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

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

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

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

Hence Proved.