What is $\tan \alpha$ if $\sin \alpha + \cos \alpha = \frac{\sqrt{3}-1}{2}$ and $\alpha \in (90^\circ,135^\circ)$ I am stuck with this problem. Any ideas on how to solve this?

The addition theorem for the sine function gives:
$$
\sin(\alpha + \pi/4) = \sin(\alpha) \cos(\pi/4) + \cos(\alpha) \sin(\pi/4)
$$
The triangle 
$$
A=(0,0), B=(1,0), C=(1,1)
$$ 
gives:
$$
\cos(\pi/4) = \sin(\pi/4) = 1/\sqrt{2}
$$
So we have
$$
\frac{\sqrt{3}-1}{2} = \sin(\alpha) + \cos(\alpha) = \sqrt{2} \sin(\alpha + \pi/4)
$$
Inverting gives
\begin{align}
\alpha &= 2\pi k + \arcsin\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right) - \pi/4 \vee \\
\alpha &= 2\pi k - \arcsin\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right) - (\pi/4 + \pi)
\end{align}
Using a computer algebra system, one finds
$$
\arcsin\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right) = \pi/12
$$
which gives
\begin{align}
\alpha &= 2\pi k + \pi/12 - \pi/4 = 2\pi k - \pi/6 = k\cdot 360^\circ - 30^\circ\vee \\
\alpha &= 2\pi k - \pi/12 - (\pi/4 + \pi) = 2\pi k - 4\pi/3 = k\cdot 360^\circ - 240^\circ
\end{align}
and then $\alpha = 120^\circ \in (90^\circ, 135^\circ)$.
which then gives $\tan 120^\circ = -\sqrt{3}$.
$$
\tan(120^\circ)
= \frac{\sin(120^\circ)}{\cos(120^\circ)}
= -\frac{\cos(30^\circ)}{\sin(30^\circ)} = -\frac{\sqrt{3}/2}{1/2}
$$
