If `cosy=xcos(a+y)` , with `cosa!=+-1` , prove that `(dy)/(dx)=(cos^2(a+y))/(sina)` .

Explanation

We have, `\cos y=x \cos (a+y) \rightarrow(1)`



Differentiate both sides w.r.t. `x`
$$
\begin{aligned}
&-\sin y \frac{d y}{d x}=\cos (a+y)-x \sin (a+y) \frac{d y}{d x} \\
&\Rightarrow \frac{d y}{d x}=\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} \\
&\quad=\frac{\cos ^{2}(a+y)}{\cos y \sin (a+y)-\sin y \cos (a+y)}=\frac{\cos ^{2}(a+y)}{\sin a}
...