The solution of the differential equation dydx=sin(x+y)tan(x+y)-1 is
cscx+y+tan(x+y)=C
x+cscx+y=C
x+tan(x+y)=C
x+secx+y=C
Explanation for the correct option
Given: dydx=sin(x+y)tan(x+y)-1
Let x+y=z
differentiating both sides
⇒1+dydx=dzdx⇒dydx=dzdx-1
⇒dzdx-1=sin(z)tan(z)-1⇒dzdx=sin(z)tan(z)⇒dzdx=sin(z)·sin(z)cos(z)⇒cos(z)dzsin2(z)=dx
Let sin(z)=t
cos(z)dz=dt
⇒∫dtt2=∫dx⇒t-2+1-2+1=x⇒-1t=x
put t=sin(z)
-1sin(z)=x
put z=x+y
⇒-cscx+y=x+C
Hence, option B is correct.