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

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(B) Given equation: $	an y \frac{dy}{dx} = \sin(x+y) + \sin(x-y)$ Using the 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 equation: $	an y \frac{dy}{dx} = \sin(x+y) + \sin(x-y)$ Using the 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$ $\frac{\sin y}{\cos y} \frac{dy}{dx} = 2 \sin x \cos y$ Rearranging the terms: $\frac{\sin y}{\cos^2 y} dy = 2 \sin x dx$ 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 $-\int \frac{dt}{t^2} = -2 \cos x + c$ $\frac{1}{t} = -2 \cos x + c$ Substituting $t = \cos y$: $\sec y = -2 \cos x + c$

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

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