A) \[\sec x-\frac{1}{2}\tan y=c\]
B) \[\log \sin (x+y)=c\]
C) \[\sec x+\tan y=c\]
D) \[\sec y+2\cos x=c\]
Correct Answer: D
Solution :
Given differential equation is \[\frac{dy}{dx}\tan y=\sin (x+y)+\sin (x-y)\] \[\Rightarrow \] \[\frac{\sin y}{{{\cos }^{2}}y}dy=2\sin x\,\,dx\] On integrating both sides, we get \[\frac{1}{\cos y}=-2\cos x+c\] \[\Rightarrow \] \[\sec y+2\cos x=c\]
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