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?
 A: Avoid squaring whenever possible as it immediately introduces extraneous root(s) 
$$\sin\alpha+\cos\alpha=\frac{\sqrt3}2-\frac12$$
$$\frac{\sin\alpha+\cos\alpha}{\sqrt2}=\sin\frac\pi4\cos\frac\pi6-\cos\frac\pi4\sin\frac\pi6$$
$$\sin\left(\alpha+\frac\pi4\right)=\sin\left(\frac\pi4-\frac\pi6\right)$$
$$\implies\alpha+\frac\pi4=n\pi+(-1)^n\left(\frac\pi4-\frac\pi6\right)$$ where $n$ is any integer
If $n$ is even $=2m$(say)$\alpha+\dfrac\pi4=2m\pi+\dfrac\pi{12}\iff\alpha=?$
If $n$ is odd, $=2m+1$(say) $\cdots$
A: Squaring both sides gives $$1+\sin 2\alpha =1-{\sqrt 3\over 2}$$
So $$\sin 2\alpha=-{\sqrt 3\over 2}$$
Find $\tan \alpha$ using the identity $$\sin 2\alpha={2\tan \alpha\over 1+\tan^2\alpha}$$
A: solve the equation
$$2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2
 \right)  \right) ^{2}}}
+{\frac {1- \left( \tan \left( x/2 \right)  \right) ^{2}}{1+ \left( 
\tan \left( x/2 \right)  \right) ^{2}}}
=\frac{\sqrt{3}-1}{2}$$ for $\tan\left(\frac{x}{2}\right)$
A: $$\sin\alpha+\cos\alpha= \frac{\sqrt3-1}2 $$
Squaring ( Squaring introduces an extraneous root, to be removed later),
$$ 1 + \sin 2 \alpha = 1- \frac{{\sqrt3}}{2} ,\; \sin 2 \alpha = \frac{-\sqrt3}{2} $$
$$ 2 \alpha = 180-60, 180 +60 $$
$$ \alpha = 60^0, 120^0  $$ 
We find that by substitution only the second solution satisfies the given condition. So,$ \alpha = 120^0 $
$$ \tan 120^0 = -\sqrt 3. $$  
A: $$\sin\alpha+\cos\alpha=\sqrt2\sin\left(\alpha+\frac{\pi}4\right)=\frac{\sqrt3-1}2\implies \alpha=\arcsin\left(\frac{\sqrt3-1}{2\sqrt2}\right)-\frac{\pi}4$$
Now:
$$\tan\alpha=\tan\left(\arcsin\left(\frac{\sqrt3-1}{2\sqrt2}\right)-\frac{\pi}4\right)=\frac{\tan\arcsin\left(\frac{\sqrt3-1}{2\sqrt2}\right)-\tan\frac{\pi}4}{1+\tan\arcsin\left(\frac{\sqrt3-1}{2\sqrt2}\right)\tan\frac{\pi}4}$$
Now:
$$\tan\arcsin\left(\frac{\sqrt3-1}{2\sqrt2}\right)=\frac{\sin\arcsin\left(\frac{\sqrt3-1}{2\sqrt2}\right)}{\cos \arcsin\left(\frac{\sqrt3-1}{2\sqrt2}\right)}=\frac{\sin\arcsin\left(\frac{\sqrt3-1}{2\sqrt2}\right)}{\sqrt{1-\sin^2 \arcsin\left(\frac{\sqrt3-1}{2\sqrt2}\right)}}=\frac{\frac{\sqrt3-1}{2\sqrt2}}{\sqrt{1-\left(\frac{\sqrt3-1}{2\sqrt2}\right)^2}}=2-\sqrt3$$
Now answer is:
$$\tan\alpha=\frac{2-\sqrt3-1}{1+2-\sqrt3}=\frac{1-\sqrt3}{3-\sqrt3}=-\frac1{\sqrt3}$$
A: 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 $\tan 120^\circ = -\sqrt{3}$.
$$
\tan(120^\circ)
= \frac{\sin(120^\circ)}{\cos(120^\circ)}
= -\frac{\cos(30^\circ)}{\sin(30^\circ)} = -\frac{\sqrt{3}/2}{1/2}
$$
A: 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
