Question

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

 

Answer 1

We have,
\[\tan y \frac{dy}{dx} = \sin \left( x + y \right) + \sin \left( x - y \right)\]
\[ \Rightarrow \tan y \frac{dy}{dx} = \sin x \cos y + \cos x \sin y + \sin x \cos y - \cos x \sin y\]
\[ \Rightarrow \tan y \frac{dy}{dx} = 2 \sin x\cos y\]
\[ \Rightarrow \frac{\tan y}{\cos y}dy = 2 \sin x dx\]
\[ \Rightarrow \tan y \sec y dy = 2 \sin x dx\]
Integrating both sides, we get
\[\int\tan y \sec y dy = 2\int\sin x dx\]
\[ \Rightarrow \sec y = - 2 \cos x + C\]
\[ \Rightarrow \sec y + 2 \cos x = C\]
\[\text{ Hence,} \sec y + 2 \cos x = \text{ C is the required solution .}\]