Question

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

Answer 1

$$\begin{gathered} \cos y=x\cos \left(a+y\right) \\ \Rightarrow -\sin y\frac{dy}{dx}=\cos \left(a+y\right)-x\sin \left(a+y\right)\frac{dy}{dx} \\ \Rightarrow -\sin y\frac{dy}{dx}+x\sin \left(a+y\right)\frac{dy}{dx}=\cos \left(a+y\right) \\ \Rightarrow \frac{dy}{dx}\left[\frac{\cos y}{\cos \left(a+y\right)}\sin \left(a+y\right)-\sin y\right]=\cos \left(a+y\right)\left[\because x=\frac{\cos y}{\cos \left(a+y\right)}\right] \\ \Rightarrow \frac{dy}{dx}\left[\frac{\cos y\sin \left(a+y\right)-\sin y\cos \left(a+y\right)}{\cos \left(a+y\right)}\right]=\cos \left(a+y\right) \\ \Rightarrow \frac{dy}{dx}\left[\frac{\sin \left(a+y-y\right)}{\cos \left(a+y\right)}\right]=\cos \left(a+y\right) \\ \Rightarrow \frac{dy}{dx}\left[\frac{\sin a}{\cos \left(a+y\right)}\right]=\cos \left(a+y\right) \\ \Rightarrow \frac{dy}{dx}=\frac{{\cos }^2\left(a+y\right)}{\sin a} \end{gathered}$$