The solution of tan y frac{dy}{dx} = sin(x+y) | vedclass

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(B) Given: $	an y \frac{dy}{dx} = \sin(x+y) + \sin(x-y)$Using the trigonometric identity $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})

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$\sec y = -2 \cos x + c$

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(B) Given: $	an y \frac{dy}{dx} = \sin(x+y) + \sin(x-y)$Using the trigonometric identity $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:$	an y \frac{dy}{dx} = 2 \sin x \cos y$Rearranging the terms to separate variables:$\frac{\sin y}{\cos^2 y} dy = 2 \sin x dx$Integrating both sides:$\int \frac{\sin y}{\cos^2 y} dy = \int 2 \sin x dx$Let $t = \cos y$,then $dt = -\sin y dy$,so $\sin y dy = -dt$.Substituting this into the integral:$-\int \frac{dt}{t^2} = -2 \cos x + c$$-(- \frac{1}{t}) = -2 \cos x + c$$\frac{1}{\cos y} = -2 \cos x + c$$\sec y = -2 \cos x + c$

The solution of $\tan y \frac{dy}{dx} = \sin(x+y) + \sin(x-y)$ is

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