A) \[\tan y+\cot x=c\]
B) \[\tan y-\cot x=c\]
C) \[\tan x-\cot y=c\]
D) \[\tan x+\cot x=c\]
Correct Answer: B
Solution :
We have, \[\frac{d\,y}{d\,x}+\frac{1+\cos 2y}{1-\cos \,2x}=0\] \[\Rightarrow \] \[\frac{dy}{dx}=-\frac{1+\cos 2y}{1-\cos 2x}=-\frac{1+2{{\cos }^{2}}y-1}{1-(1-2{{\sin }^{2}}x)}\] \[\Rightarrow \] \[\frac{dy}{dx}=-\frac{2{{\cos }^{2}}y}{2{{\sin }^{2}}x}\] \[\Rightarrow \] \[\int{{{\sec }^{2}}y\,dy=-}\int{\cos e{{c}^{2}}x\,dx}\] \[\Rightarrow \] \[\tan y-\cot \,x=c\]
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