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?

Let $\sin A=\dfrac{\sqrt3}2,\cos A=-\dfrac12$
Clearly, $A=120^\circ$ is a solution
So, we can write $\sin\alpha-\sin A=\cos A-\cos\alpha$
Using Prosthaphaeresis Formulas, $$2\sin\frac{\alpha-A}2\cos\frac{\alpha+A}2=2\sin\frac{\alpha-A}2\sin\frac{\alpha+A}2$$
$$\iff\sin\frac{\alpha-A}2\left[\cos\frac{\alpha+A}2-\sin\frac{\alpha+A}2\right]=0$$
If $\sin\dfrac{\alpha-A}2=0,\dfrac{\alpha-A}2=180^\circ n\iff \alpha=A+360^\circ n$ where $n$ is any integer
Else $\tan\dfrac{\alpha+A}2=1=\tan45^\circ$
$\implies\dfrac{\alpha+A}2=m180^\circ+45^\circ\iff\alpha=360^\circ m+90^\circ-A$ where $m$ is any integer
