The problem is as follows
Use rectangular coordinates for the disk $\Bbb D = \{(x,y):x^2+y^2\le 1\}$, and define a function $u(x,y)=\arctan(\frac{y}{1+x})$
(a)Show that $u$ is constant on straight line segments through $(-1,0)$ hence evaluate $f(\theta)=\lim_{r \to 1-}u(r\cos\theta, r\cos\theta$)
Attempt : (a) suppose $y=m(x+1)$
$$ u(x)=\arctan\left(\frac{m(x+1)}{x+1}\right)=\arctan(m)$$
so $u$ is constant
Now consider $$f(\theta)=\lim_{r \to 1-}u(r \cos\theta, r \cos\theta) = \lim_{r \to 1-} \arctan\left(\frac{r \sin\theta}{1 + r \cos\theta} \right) = \arctan \left(\frac{\sin\theta}{1+\cos\theta}\right)$$
for $-\pi\lt \theta \lt \pi$
$$ \arctan\left(\frac{\sin\theta}{1+\cos\theta}\right) = \arctan\left(\frac{2\sin\frac{\theta}{2} \cos\frac{\theta}{2}}{2 \cos^2\frac{\theta}{2}}\right) = \arctan\left(\tan\left(\frac{\theta}{2} \right)\right)=\frac{\theta}{2}$$ $QED$
Is this correct? But Why do I show $u$ is constant to evaluate $f(\theta)$?