A) \[\text{cosec}\left( x+y \right)+\text{tan}\left( x+y \right)=x+c\]
B) \[x+\text{cosec}\left( x+y \right)=c\]
C) \[x+\text{tan}\left( x+y \right)=c\]
D) \[x+\text{sec}\left( x+y \right)=c\]
Correct Answer: B
Solution :
Given, \[\frac{dy}{dx}=\sin \,(x+y)\tan (x+y)-1\] Put \[x+y=z\] \[\Rightarrow \] \[1+\frac{dy}{dx}=\frac{dz}{dx}\] \[\therefore \] \[\frac{dz}{dx}-1=\sin z\tan z-1\] \[\Rightarrow \] \[\int{\frac{\cos z}{{{\sin }^{2}}z}}\,dz=\int{dx}\] Put \[\sin z=t\]\[\Rightarrow \]\[\cos z\,dz=dt\] \[\therefore \] \[\int{\frac{1}{{{t}^{2}}}\text{dt}}=x-c\] \[\Rightarrow \] \[-\frac{1}{t}=x-c\] \[\Rightarrow \] \[-\text{cosec}\,z\,=x-c\] \[\Rightarrow \] \[x+\text{cosec}\,(x+y)=c\]
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