Question
Solve the following differential equations:$\tan\text{y}\frac{\text{dy}}{\text{dx}}=\sin(\text{x}+\text{y})+\sin(\text{x}-\text{y})$

Answer

$\tan\text{y}\frac{\text{dy}}{\text{dx}}=\sin(\text{x}+\text{y})+\sin(\text{x}-\text{y})$
$\tan\text{y}\frac{\text{dy}}{\text{dx}}=2\sin\Big\{\frac{(\text{x + y})+(\text{x}-\text{y})}{2}\Big\}\cos\Big\{\frac{(\text{x + y})-(\text{x}-\text{y})}{2}\Big\}$
$=2\sin\Big(\frac{\text{x + y + x}-\text{y}}{2}\Big)\cos\Big(\frac{\text{x + y}-\text{ x}+\text{y}}{2}\Big)$
$\tan\text{y}\frac{\text{dy}}{\text{dx}}=2\sin\text{x}\cos\text{y}$
$\frac{\tan\text{y}}{\cos\text{y}}\text{dy}=2\sin\text{x dx}$
$\int\sec\text{y}\tan\text{y dy}=2\int\sin\text{x dx}$
$\sec\text{y}=-2\cos\text{x + C}$
$\sec\text{y}+2\cos\text{x = C}$

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