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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

@Toky,

There is nothing wrong with your methods. Both are correct. You are getting two different representations for the same set of solutions. According to the second solution $$ \dots -\frac{\pi}{6}, \frac{\pi}{6}, \frac{3\pi}{6}, \frac{5\pi}{6},\dots $$ Your first solutions gives you half of these if you choose the $+$ sign, and gives you the other half if you choose the $-$ sign.

Hope this helps.