A) \[\sin y+\cos x=c\]
B) \[\sin y-\cos x=c\]
C) \[\sin y.\cos x=c\]
D) \[\sin y=c\cos x\]
Correct Answer: D
Solution :
\[\cos x\cos y\frac{dy}{dx}=-\sin x\sin y\] Þ \[\frac{\cos y}{\sin y}dy=-\frac{\sin x}{\cos x}dx\] Þ \[\cot ydy=-\tan xdx\] On integrating, we get \[\log \sin y=\log \cos x+\log c\]Þ\[\sin y=c\cos x\].You need to login to perform this action.
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