Solution of differential equation $\frac{dy}{dx}=cos(x+y+3)$ is :

$y=2tan^{-1}(x+c)-x-3$

The correct answer is Option (3) → $y=2tan^{-1}(x+c)-x-3$

$\frac{dy}{dx}=\cos(x+y+3)$

let $z=x+y+3$

so $\frac{dz}{dx}=1+\frac{dy}{dx}$

$\frac{dz}{dx}-1=\cos z$

so $\int\frac{1}{1+\cos^2z}dz=\int dx$

$=\int\frac{1-\cos z}{1+\cos^2z}dz=\int dx$

$=\int\frac{1}{\sin^2z}-\frac{\cos z}{\sin^2z}dz=\int dx$

$⇒\int cosec^z-cosec\,z\cot z\,dz=\int dz$

$≡cosec\,z-\cot z=x+c$

$\frac{1-\cos z}{\sin z}=x+c$

$=\frac{2\sin^2\frac{z}{2}}{2\sin\frac{z}{2}\cos\frac{z}{2}}=x+c$

$\tan\frac{z}{2}=x+c$

$z=2\tan^{-1}(x+c)$

$x+y+3=2\tan^{-1}(x+c)$

$y=2\tan^{-1}(x+c)-x-3$