A) \[{{e}^{\cos x}}\tan \frac{y}{2}=c\]
B) \[{{e}^{\cos x}}\tan y=c\]
C) \[\cos x\tan y=c\]
D) \[\cos x\sin y=c\]
Correct Answer: A
Solution :
\[\int_{{}}^{{}}{\frac{dy}{\sin y}}=\int_{{}}^{{}}{\sin xdx}\]Þ\[\log \tan \frac{y}{2}=-\cos x+c\] Þ \[\tan \frac{y}{2}={{e}^{-\cos x+c}}\]Þ\[{{e}^{\cos x}}\tan \frac{y}{2}={{e}^{C}}=c\].
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