Question

TanΦ+tan2Φ+√3 tanΦtan2Φ=√3

Answer 1

➡HERE IS YOUR ANSWER⬇

$tan \alpha + tan2 \alpha + \sqrt{3} tan \alpha tan2 \alpha = \sqrt{3} \\ \\ or \: \: tan \alpha + tan2 \alpha = \sqrt{3} (1 - tan \alpha \: tan2 \alpha ) \\ \\ or \: \: \frac{tan \alpha + tan2 \alpha }{1 - tan \alpha \: tan2 \alpha } = \sqrt{3} \\ \\ or \: \: tan( \alpha + 2 \alpha ) = \sqrt{3} \\ \\ or \: \: tan3 \alpha = tan \frac{\pi}{3} \\ \\ or \: \: 3 \alpha = \frac{\pi}{3} \\ \\ hence \: \: \alpha = \frac{\pi}{9}$

Rule to remember :

$tan( \alpha + \beta ) = \frac{tan \alpha + tan \beta }{1 - tan \alpha \: tan \beta }$

⬆HOPE THIS HELPS YOU⬅