Question

tan y$\frac{dy}{dx}$ = sin (x + y) + sin (x − y)

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \tan y\frac{dy}{dx}=\sin \left(x+y\right)+\sin \left(x-y\right) \\ \Rightarrow \tan y\frac{dy}{dx}=\sin x\cos y+\cos x\sin y+\sin x\cos y-\cos x\sin y \\ \Rightarrow \tan y\frac{dy}{dx}=2\sin x\cos y \\ \Rightarrow \frac{\tan y}{\cos y}dy=2\sin xdx \\ \Rightarrow \tan y\sec ydy=2\sin xdx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \tan y\sec ydy=2\int \sin xdx \\ \Rightarrow \sec y=-2\cos x+C \\ \Rightarrow \sec y+2\cos x=C \\ \mathrm{Hence},\sec y+2\cos x=C\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$