Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},\cos y=x\cos \left(a+y\right) \\ \mathrm{Differentiating}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get}, \\ \frac{d}{dx}\left(\cos y\right)=\frac{d}{dx}\left\{x\cos \left(a+y\right)\right\} \\ \Rightarrow -\sin y\frac{dy}{dx}=\cos \left(a+y\right)\frac{d}{dx}\left(x\right)+x\frac{d}{dx}\cos \left(a+y\right) \\ \Rightarrow -\sin y\frac{dy}{dx}=\cos \left(a+y\right)+x\left[-\sin \left(a+y\right)\right]\frac{dy}{dx} \\ \Rightarrow \left[x\sin \left(a+y\right)-\sin y\right]\frac{dy}{dx}=\cos \left(a+y\right) \\ \Rightarrow \left[\frac{\cos y}{\cos \left(a+y\right)}\sin \left(a+y\right)-\sin y\right]\frac{dy}{dx}=\cos \left(a+y\right)\left[\because \cos y=x\cos \left(a+y\right)\Rightarrow x=\frac{\cos y}{\cos \left(a+y\right)}\right] \\ \Rightarrow \left[\cos y\sin \left(a+y\right)-\sin y\cos \left(a+y\right)\right]\frac{dy}{dx}={\cos }^2\left(a+y\right) \\ \Rightarrow \sin \left(a+y-y\right)\frac{dy}{dx}={\cos }^2\left(a+y\right) \\ \Rightarrow \frac{dy}{dx}=\frac{{\cos }^2\left(a+y\right)}{\sin a} \end{gathered}$$