If $\frac{dy}{dx}=\cos (x+y),y\left(\frac{\pi }{2}\right)=0$ then $y\left(0\right)=$

$x+y=z\Rightarrow 1+\frac{dy}{dx}=\frac{dz}{dx},\frac{dz}{dx}=1+\cos z$

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

$\Rightarrow x=\tan \frac{z}{2}+c\text{ or }x=\tan \left(\frac{x+y}{2}\right)+c$

$$\begin{gathered} y\left(\frac{\pi }{2}\right)=0 \\ \Rightarrow \frac{\pi }{2}=1+c \\ \Rightarrow c=\frac{\pi }{2}-1 \end{gathered}$$

$$\begin{gathered} x=\tan \left(\frac{x+y}{2}\right)+\frac{\pi }{2}-1 \\ \Rightarrow y=-x+2{\tan }^{-1}\left(x+1-\frac{\pi }{2}\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow y\left(0\right)=2{\tan }^{-1}\left(1-\frac{\pi }{2}\right) \\ \Rightarrow y\left(0\right)=-2{\tan }^{-1}\left(\frac{\pi }{2}-1\right) \end{gathered}$$