Q:

If $\tan 2\alpha \cdot \tan \alpha = 1$, then what is $\alpha$? Different methods give different answers.

If $\tan 2\alpha\cdot\tan \alpha = 1$, then what is $\alpha$?

I tried two methods but got two different answers.
Method 1:
$$\begin{align}
\tan 2\alpha\cdot\tan \alpha = 1
&\implies
\frac{2\tan \alpha}{1 - \tan ^2 \alpha}\;\tan \alpha = 1 \tag{1a}\\[6pt]
&\implies 2\tan ^2\alpha = 1 - \tan ^2\alpha \tag{1b}\\[6pt]
&\implies \tan ^2 \alpha = \frac{1}{3} \tag{1c}\\[6pt]
&\implies \tan \alpha = \pm\frac{1}{\sqrt 3} \tag{1d}\\[6pt]
&\implies \tan \alpha = \tan\left(\pm\frac{\pi}{6}\right) \tag{1e}\\[6pt]
&\implies \alpha = n\pi \pm \frac{\pi}{6} \;\text{where}\; n \in \mathbb{Z} \tag{1f}
\end{align}$$
Method 2:
$$\begin{align}
\tan 2\alpha \cdot \tan \alpha = 1
&\implies \tan 2\alpha = \frac{1}{\tan \alpha} \tag{2a}\\[6pt]
&\implies \tan 2\alpha = \cot \alpha \tag{2b}\\[6pt]
&\implies \tan 2\alpha = \tan\left(\frac{\pi}{2} - \alpha\right) \tag{2c}\\[6pt]
&\implies 2\alpha = n\pi + \frac{\pi}{2} - \alpha \text{?} \tag{2d}\\[6pt]
&\implies \alpha = \frac{1}{3}\left(n\pi + \frac{\pi}{2}\right)\;\text{where}\; n \in \mathbb{Z} \tag{2e}
\end{align}$$
Which one is correct? Is there any mistake in the above solutions?

A:

Note carefully for future reference that in solving equations you need to use $\Leftrightarrow$ not $\Rightarrow$, or to do something equivalent.
Both your answers are correct as far as they go, but incomplete.
In your first method, for $\alpha=n\pi+\frac\pi6$ we check that
$$\tan2\alpha\tan\alpha=\sqrt3\frac1{\sqrt3}=1\ ,$$
so this is a correct solution, while for $\alpha=n\pi-\frac\pi6$ we have
$$\tan2\alpha\tan\alpha=(-\sqrt3)(-\frac1{\sqrt3})=1\ ,$$
so this is also a correct solution.
For your second solution there are three cases: $n=3k$ or $n=3k+1$ or $n=3k+2$.  The first gives
$$\tan2\alpha\tan\alpha=\sqrt3\frac1{\sqrt3}=1\ ,$$
so this is a correct solution.  The second gives $\alpha=k\pi+\frac\pi2$, so $\tan\alpha$ is undefined and this must be ruled out.  The third gives
$$\tan2\alpha\tan\alpha=\sqrt3\frac1{\sqrt3}=1\ ,$$
so this is also a correct solution.  Therefore your second method should give an answer
$$\alpha=\tfrac13(n\pi+\tfrac\pi2)\ ,\quad\hbox{where $n=3k$ or $n=3k+2$}.$$
The actual numbers obtained in this solution are then the same as in your first method.
Always check your answers if you start with an equation and derive (potential) solutions.

A:

Your first method is correct.
In the second method, notice that the equation
$$\tan 2\alpha = \frac{1}{\tan\alpha}$$
is not valid when $\alpha = \frac{\pi}{2} + n\pi, n \in \mathbb{Z}$ or when $\alpha = \frac{\pi}{4} + \frac{n\pi}{2}, n \in \mathbb{Z}$ or when $\alpha = n\pi, n \in \mathbb{Z}$.  Therefore, in your solution 
$$\alpha = \frac{\pi}{6} + \frac{n\pi}{3}, n \in \mathbb{Z}$$
$n \neq 3k + 1$, $k \in \mathbb{Z}$ since that would imply
$$\alpha = \frac{\pi}{6} + k\pi + \frac{\pi}{3} = \frac{\pi}{2} + k\pi, k \in \mathbb{Z}$$
which is not valid.  
If we replace $n$ by $3k$, we obtain
$$\alpha = \frac{\pi}{6} + k\pi, k \in \mathbb{Z}$$
If we replace $n$ by $3k - 1$, we obtain
$$\alpha = \frac{\pi}{6} + k\pi - \frac{\pi}{3} = -\frac{\pi}{6} + k\pi, k \in \mathbb{Z}$$
which agrees with your first solution.

