A) \[\cos ec(x+y)+\tan (x+y)=x+c\]
B) \[x+\cos ec(x+y)=c\]
C) \[x+\tan (x+y)=c\]
D) \[x+\sec (x+y)=c\]
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 \[\sin z=t\] \[\Rightarrow \] \[\cos z\,\,dz=dt\] \[\therefore \] \[\int{\frac{1}{{{t}^{2}}}dt}=x-c\] \[\Rightarrow \] \[-\frac{1}{t}=x-c\] \[\Rightarrow \] \[-\cos \text{ec}z=x-c\] \[\Rightarrow \] \[x+\cos \text{ec}(x+y)=c\]
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