QUESTION

# Prove the given trigonometric equation $\tan A + \tan ({60^o} + A) + \tan ({120^o} + A) = 3\tan 3A.$

(i)$\tan ({180^o} - A) = - \tan A$
(ii)$\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
$L.H.S$
$= \tan A + \tan ({60^o} + A) + \tan ({120^o} + A) \\ = \tan A + \tan ({60^o} + A) + \tan ({180^o} - ({60^o} - A)) \\ = \tan A + \tan ({60^o} + A) - \tan ({60^o} - A) \\ = \tan A + \dfrac{{\tan {{60}^o} + \tan A}}{{1 - \tan {{60}^o}\tan A}} - \dfrac{{\tan {{60}^o} - \tan A}}{{1 + \tan {{60}^o}\tan A}} \\ = \tan A + \dfrac{{\sqrt 3 + \tan A}}{{1 - \sqrt 3 \tan A}} - \dfrac{{\sqrt 3 - \tan A}}{{1 + \sqrt 3 \tan A}} \\ = \tan A + \dfrac{{(\sqrt 3 + \tan A)(1 + \sqrt 3 \tan A) - (\sqrt 3 - \tan A)(1 - \sqrt 3 \tan A)}}{{(1 - \sqrt 3 \tan A)(1 + \sqrt 3 \tan A)}} \\ = \tan A + \left[ {\dfrac{{\sqrt 3 + 3\tan A + \tan A + \sqrt 3 {{\tan }^2}A - \sqrt 3 + 3\tan A + \tan A - \sqrt 3 {{\tan }^2}A}}{{{1^2} - {{(\sqrt 3 \tan A)}^2}}}} \right] \\ = \tan A + \left[ {\dfrac{{8\tan A}}{{1 - 3{{\tan }^2}A}}} \right] \\ = \left[ {\dfrac{{\tan A - 3{{\tan }^3}A + 8\tan A}}{{1 - 3{{\tan }^2}A}}} \right] \\ = \dfrac{{9\tan A - 3{{\tan }^3}A}}{{1 - 3{{\tan }^2}A}} \\$
$= 3\left( {\dfrac{{3\tan A - {{\tan }^3}A}}{{1 - 3{{\tan }^2}A}}} \right) \\ = 3\tan 3A \\$
$= R.H.S$
Note: To solve this type of question, we must remember some basic trigonometric identity formulae. One also must know the value of $\tan {60^o}$. One basic identity used is this problem is
$\tan 3A = \left( {\dfrac{{3\tan A - {{\tan }^3}A}}{{1 - 3{{\tan }^2}A}}} \right)$