The solution of $cos (x+y) dy = dx,$  is

$y = tan\left(\frac{x+y}{2}\right) +C$

The correct answer is option (1) : $y = tan\left(\frac{x+y}{2}\right) +C$

We have, $cos (x + y ) dy = dx$

Putting $ x+ y = v$ and $1+\frac{dy}{dx} =\frac{dv}{dx},$ we  get

$cos\, v \left(\frac{dv}{dx} - 1\right ) = 1 $

$⇒\frac{dv}{dx} = \frac{1+cosv}{cosv}$

$⇒\frac{cosv}{1+cosv}dv=dx$

$⇒\left(1-\frac{1}{1+cosv}\right) dv= dx$

$⇒∫\left(1-\frac{1}{2}sec^2 \frac{v}{2} \right) dv = ∫dx$

$⇒v-tan\frac{v}{2}=x+C$

$⇒x + y -  tan \left(\frac{x+y}{2} \right) = x+ C$

$⇒y = tan \frac{x+y}{2} + C$