A) \[\tan y+\cot x=c\]
B) \[\tan y\cot x=c\]
C) \[\tan y-\cot x=c\]
D) None of these
Correct Answer: C
Solution :
\[\frac{dy}{dx}=-\frac{1+\cos 2y}{1-\cos 2x}\] Þ \[\frac{dy}{dx}=-\frac{2{{\cos }^{2}}y}{2{{\sin }^{2}}x}\] Þ \[{{\sec }^{2}}ydy=-\text{cose}{{\text{c}}^{2}}xdx\] On integrating both sides, we get \[\tan y=\cot x+c\] Þ \[\tan y-\cot x=c\].
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