$\tan y \frac{d y}{d x}=\sin (x+y)+\sin (x-y)$
$\tan y \frac{d y}{d x}=2 \cdot \sin \left(\frac{2 x}{2}\right) \cdot \cos \left(\frac{2 y}{2}\right)$
$\left[\because \sin C+\sin D=2 \sin \left(\frac{C+D}{2}\right) \cdot \cos \left(\frac{C-D}{2}\right)\right]$
$\Rightarrow \quad \tan y \frac{d y}{d x}=2 \sin x \cdot \cos y$
$\Rightarrow \quad \frac{\sin y}{\cos y} \frac{d y}{d x}=2 \sin x \cos y$
On integrating
$\Rightarrow \quad \int \frac{\sin y}{\cos ^2 y} d y=\int 2 \sin x d x$
$\left[\begin{array}{l}\text { Let } t=\cos y \\ \frac{d t}{d y}=-\sin y \\ -d t=\sin y d y\end{array}\right]$
$\Rightarrow \quad-\int \frac{d t}{t^2}=2(-\cos x)+c$
$\Rightarrow \quad-\left(-\frac{1}{t}\right)=-2 \cos x+c$
$\Rightarrow \quad \frac{1}{\cos y}=-2 \cos x+c$
$\Rightarrow \quad \sec y=-2 \cos x+c$