If $\tan(\alpha)=2-\sqrt{3},$ find the value of the acute angle $\alpha$ I rewrote it as $$\tan\left(2\cdot\frac{\alpha}{2}\right)=\frac{2\tan(\frac{\alpha}{2})}{1-\tan^2(\frac{\alpha}{2})}=2-\sqrt{3} \ \Rightarrow \ 2(1-\tan^2(\frac{\alpha}{2})-\sqrt{3}(2\tan(\frac{\alpha}{2}).$$ Obviously, it didn't bring the answer. My other attempts also were untrue. I would like somebody to hint at other method(s).
 A: hint
Put $$\beta=\frac {\pi}{2}-\alpha $$
then
$$\tan (\beta)=\frac {1}{2-\sqrt {3}}=2+\sqrt {3}$$
and
$$\tan(\beta)-\tan (\alpha)=2\sqrt {3}$$
hence
$$\tan (\beta-\alpha)=\sqrt {3}=\tan(\frac {\pi}{3}) $$
and
$$2\alpha=\frac {\pi}{2}-\frac {\pi}{3}=\frac {\pi}{6} $$
Done!
A: You should pay more attention to your hints.
$\tan \alpha = 2-\sqrt{3}$
Since we are given that $\alpha$ is acute and $0 < \tan \alpha < 1$, then 
$0 < \alpha < \frac{\pi}{4}$ and $0 < 2\alpha < \frac{\pi}{2}$.
Since \begin{align}
   \tan 2\alpha
   &= \dfrac{2\tan \alpha}{1 - \tan^2 \alpha} \\
   &= \dfrac{4-2\sqrt 3}{1 - (7-4 \sqrt 3)} \\
   &= \dfrac{4-2\sqrt 3}{-6 + 4 \sqrt 3} \\
   &= \dfrac{2(2 - \sqrt 3)}{2\sqrt 3(2 - \sqrt 3)} \\
   &= \dfrac{1}{\sqrt 3} \\
   &= \tan \frac{\pi}{6}
\end{align}
Then $\alpha = \dfrac{\pi}{12}$
A: Just it's better to know that $\alpha=15^{\circ}$ and the rest is smooth: 
$$\tan15^{\circ}=\tan(45^{\circ}-30^{\circ})=\frac{1-\frac{1}{\sqrt3}}{1+\frac{1}{\sqrt3}}=\frac{\sqrt3-1}{\sqrt3+1}=\frac{(\sqrt3-1)^2}{2}=2-\sqrt3$$
and since you want an acute angle, we are done!
A: $$\csc2A-\cot2A=\dfrac{1-\cos2A}{\sin2A}=\cdots=\tan A$$
Here $2A\equiv60^\circ\pmod{360^\circ}$
