1) The solution of the differential equation dydx=1−cos(y−x)cot(y−x) is A) x tan (y-x)=c B) x= tan(y-x)+c C) x= sec(y-x)+c D) x+ sec (y-x)=c Answer: Option DExplanation:We have dydx=1−cos(y−x)cot(y−x) Put y−x=v⇒dydx=1+dvdx ∴ 1+dvdx=1−cosvcotv ⇒ dvdx=−cos2vsinv ⇒ −∫sinvcos2vdv=∫dx⇒−∫secvtanvdv=dx ⇒−secv=x+c⇒x+sec(y−x)=c