Question

If $\tan \theta =a\tan \frac{\theta }{3}$, express$\tan \frac{\theta }{3}$in terms of a.

Short answer

On expressing $\tan \frac{\theta }{3}$ in terms of a, we get $0,\sqrt{\frac{a-3}{3a-1}},-\sqrt{\frac{a-3}{3a-1}}$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$\tan \theta =a\tan \frac{\theta }{3}$

We know localid="1646426304523" $\tan \left(3\theta \right)=\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$. Then,

localid="1646427644279" role="math" $$\begin{gathered} \frac{3\tan \frac{\theta }{3}-{\tan }^3\frac{\theta }{3}}{1-3{\tan }^2\frac{\theta }{3}}=a\tan \frac{\theta }{3} \\ 3\tan \frac{\theta }{3}-{\tan }^3\frac{\theta }{3}=a\tan \frac{\theta }{3}\left(1-3{\tan }^2\frac{\theta }{3}\right) \end{gathered}$$

Step 2. Simply the equation.

Continuing the above equation,

$$\begin{gathered} 3\tan \frac{\theta }{3}-{\tan }^3\frac{\theta }{3}=a\tan \frac{\theta }{3}-3a{\tan }^3\frac{\theta }{3} \\ \left(3a-1\right){\tan }^3\frac{\theta }{3}+\left(3-a\right)\tan \frac{\theta }{3}=0 \\ \tan \frac{\theta }{3}=0,\pm \sqrt{\frac{a-3}{3a-1}} \end{gathered}$$