A) \[sec\text{ }x-1/2\text{ }tan\text{ }y=c\]
B) \[log\text{ }sin\text{ (}x+y)=c\]
C) \[sec\text{ }x+tan\text{ }y=c\]
D) \[sec\text{ }y+2\text{ }cos\text{ }x=c\]
Correct Answer: D
Solution :
Given, differential equation is \[\left( \frac{dy}{dx} \right)\tan y=\sin (x+y)+\sin (x-y)\] \[\Rightarrow \] \[\left( \frac{dy}{dx} \right)\tan y=2\sin x\cos y\] \[\Rightarrow \] \[\frac{\sin y}{{{\cos }^{2}}y}dy=2\sin xdx\] 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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