Question

If $\cos y=x\cos (a+y)$, with $\cos a\ne \pm 1$, prove that $\frac{dy}{dx}=\frac{{\cos }^2(a+y)}{\sin a}$

Answer 1

We have, $\cos y=x\cos (a+y)\to \left(1\right)$
Differentiate both sides w.r.t. $x$
$-\sin y\frac{dy}{dx}=\cos (a+y)-x\sin (a+y)\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{\cos (a+y)}{x\sin (a+y)-\sin y}=\frac{\cos (a+y)}{\frac{\cos y}{\cos (a+y)}\sin (a+y)-\sin y}$
$=\frac{{\cos }^2(a+y)}{\cos y\sin (a+y)-\sin y\cos (a+y)}=\frac{{\cos }^2(a+y)}{\sin a}$
Since $\sin (A-B)=\sin A\cos B-\sin B\cos A$