<s>[INST] Different ways gives different results - solving $\tan 2a = \sqrt 3 $ Different ways gives different results - solving $\tan 2a = \sqrt 3$
Case 1). 
$$ \tan 2a = \sqrt 3 =\tan(\frac{\pi}{3}) $$
$$2a = n\pi + \frac{\pi}{3} $$
$$a = \frac{n\pi}{2} + \frac{\pi}{6},\qquad   n \in\mathbb{Z}   $$
Case 2) 
$$ \dfrac{2\tan a}{1-\tan^2 a}  = \sqrt 3$$
Solving above equation gives $ \tan a =  \dfrac{1}{\sqrt 3}$ or $-\sqrt 3 $
Thus    either 
$$ a = n\pi+ \frac{\pi}{6},\qquad   n \in \mathbb{Z} $$
Or
$$    a = m\pi - \frac{\pi}{3},\qquad     m\in\mathbb{Z} $$
Or either I did something wrong or both these results are same. 
If it's the later, I tried a lot to convert one form to other by adding the case 2 results. But that doesn't work.
 [/INST] If you are in doubt, enumerate the solutions.
Case 1:
$$\cdots-\frac{2\pi}6,\frac\pi6,\frac{4\pi}6,\frac{7\pi}6,\frac{10\pi}6,\frac{13\pi}6,\cdots$$
Case 2:
$$\cdots\frac\pi6,\frac{7\pi}6,\frac{13\pi}6,\cdots$$
and
$$\cdots-\frac{2\pi}6,\frac{4\pi}6,\frac{10\pi}6,\cdots$$
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