Question

Prove that : tanA + tan(A + π/3) + tan(A + 2π/3) = 3

Answer 1

$$\begin{array}{l} \tan A + \tan \left(A + \frac {\pi}{3}\right) + \tan \left(A + \frac {2 \pi}{3}\right) = \tan A + \tan \left(A + \frac {\pi}{3}\right) + \tan \left(\pi - \left(\frac {\pi}{3} - A\right)\right) = \\ = \tan A + \tan \left(A + \frac {\pi}{3}\right) - \tan \left(\frac {\pi}{3} - A\right) = \tan A + \tan \left(A + \frac {\pi}{3}\right) + \tan \left(A - \frac {\pi}{3}\right) = \\ = \tan A + \frac {\tan A + \tan \frac {\pi}{3}}{1 - \tan A \tan \frac {\pi}{3}} + \frac {\tan A - \tan \frac {\pi}{3}}{1 + \tan A \tan \frac {\pi}{3}} = \\ = \tan A + \frac {\tan A + \sqrt {3}}{1 - \sqrt {3} \tan A} + \frac {\tan A - \sqrt {3}}{1 + \sqrt {3} \tan A} \\ = \tan A + \frac {\tan A + \sqrt {3} + \sqrt {3} \tan^ {2} A + 3 \tan A + \tan A - \sqrt {3} - \sqrt {3} \tan^ {2} A + 3 \tan A}{1 - 3 \tan^ {2} A} = \\ = \tan A + \frac {\tan A + 3 \tan A + \tan A + 3 \tan A}{1 - 3 \tan^ {2} A} = \tan A + \frac {8 \tan A}{1 - 3 \tan^ {2} A} = \tan A \left(1 + \frac {8}{1 - 3 \tan^ {2} A}\right) = \\ = \tan A \left(\frac {1 - 3 \tan^ {2} A + 8}{1 - 3 \tan^ {2} A}\right) = 3 \left(\frac {3 \tan A - \tan^ {3} A}{1 - 3 \tan^ {2} A}\right) = 3 \tan (3 A) \neq 3 \\ \end{array}$$

I've verify this result in Mathcad:

$$\tan (x) + \tan \left(x + \frac {\pi}{3}\right) + \tan \left(x + \frac {2 \pi}{3}\right)$$

Simplify :=

3-tan(3-x)

It's true!

So, $\tan A + \tan (A + \frac{\pi}{3}) + \tan (A + \frac{2\pi}{3}) \neq 3$

$$\tan A + \frac {\tan A + \tan \frac {\pi}{3}}{1 - \tan A \tan \frac {\pi}{3}} + \frac {\tan A + \tan \frac {2 \pi}{3}}{1 - \tan A \tan \frac {2 \pi}{3}} = = \tan A + \frac {\tan A + \sqrt {3}}{1 - \sqrt {3} \tan A} + \frac {\tan A + \tan \frac {2 \pi}{3}}{1 - \tan A \tan \frac {2 \pi}{3}}$$