Question

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

Answer 1

Hint: Here the trigonometric ratio tangent is given in the form of sum of two angles, So we use the trigonometric formulas:
(i)$\tan ({180^o} - A) = - \tan A$
(ii)$\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$

Complete step-by-step solution -
$L.H.S$
$ = \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}} \\
$
Taking 3 common,
$
   = 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
$\tan 3A = \left( {\dfrac{{3\tan A - {{\tan }^3}A}}{{1 - 3{{\tan }^2}A}}} \right)$