A) \[\tan (x+y)+\sec (x+y)=x+c\]
B) \[\tan (x+y)-\sec (x+y)=x+c\]
C) \[\tan (x+y)+\sec (x+y)=x+c=0\]
D) None of these
Correct Answer: B
Solution :
Here \[\frac{dy}{dx}=\sin \,(x+y)\] Now put \[x+y=v\]and \[\frac{dy}{dx}=\frac{dv}{dx}-1\] Therefore \[\frac{dy}{dx}=\sin (x+y)\] reduces to \[\frac{dv}{1+\sin v}=dx\] Now on integrating both the sides, we get \[\int{\frac{1-\sin v}{1-{{\sin }^{2}}v}}\,\,dv=\int{dx}\] \[\Rightarrow \,\,\tan v-\sec v=x+c\] or \[\tan \,\,\left( x+y \right)-\sec \,\,\left( x+y \right)=x+c.\]
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