Question
$\tan \mathrm{A}+\tan \left(60^{\circ}+\mathrm{A}\right)+\tan \left(120^{\circ}+\mathrm{A}\right)=3 \tan 3 \mathrm{~A}$
✓
Answer
L.H.S.
$\begin{aligned}
& =\tan \mathrm{A}+\tan \left(60^{\circ}+\mathrm{A}\right)+\tan \left(120^{\circ}+\mathrm{A}\right) \\
& =\tan A+\frac{\tan 60^{\circ}+\tan A}{1-\tan 60^{\circ} \tan A}+\frac{\tan 120^{\circ}+\tan A}{1-\tan 120^{\circ} \tan A} \\
& =\tan A+\frac{\sqrt{3}+\tan A}{1-\sqrt{3} \tan A}+\frac{-\tan 60^{\circ}+\tan A}{1-\left(-\tan 60^{\circ}\right) \tan A} \\
& \cdots\left[\begin{array}{rl}
\because \tan 120^{\circ} & =\tan \left(180^{\circ}-60^{\circ}\right) \\
& =-\tan 60^{\circ}
\end{array}\right] \\
& =\tan A+\frac{\sqrt{3}+\tan A}{1-\sqrt{3} \tan A}-\frac{\sqrt{3}-\tan A}{1+\sqrt{3} \tan A} \\
& =\tan \mathrm{A}\\
& +\frac{\sqrt{3}+3 \tan A+\tan A+\sqrt{3} \tan ^2 A-\sqrt{3}+\tan A+3 \tan A-\sqrt{3} \tan ^2 A}{(1-\sqrt{3} \tan A)(1+\sqrt{3} \tan A)} \\
& =\tan A+\frac{8 \tan A}{1-3 \tan ^2 A} \\
& =\frac{\tan A-3 \tan ^3 \mathrm{~A}+8 \tan \mathrm{A}}{1-3 \tan ^2 \mathrm{~A}} \\
& =\frac{9 \tan A-3 \tan ^3 \mathrm{~A}}{1-3 \tan ^2 \mathrm{~A}} \\
& =3\left(\frac{3 \tan A-\tan ^3 \mathrm{~A}}{1-3 \tan ^2 \mathrm{~A}}\right) \\
& =3 \tan 3 \mathrm{~A} \\
& =\text { R.H.S. } \\
\end{aligned}$
$\begin{aligned}
& =\tan \mathrm{A}+\tan \left(60^{\circ}+\mathrm{A}\right)+\tan \left(120^{\circ}+\mathrm{A}\right) \\
& =\tan A+\frac{\tan 60^{\circ}+\tan A}{1-\tan 60^{\circ} \tan A}+\frac{\tan 120^{\circ}+\tan A}{1-\tan 120^{\circ} \tan A} \\
& =\tan A+\frac{\sqrt{3}+\tan A}{1-\sqrt{3} \tan A}+\frac{-\tan 60^{\circ}+\tan A}{1-\left(-\tan 60^{\circ}\right) \tan A} \\
& \cdots\left[\begin{array}{rl}
\because \tan 120^{\circ} & =\tan \left(180^{\circ}-60^{\circ}\right) \\
& =-\tan 60^{\circ}
\end{array}\right] \\
& =\tan A+\frac{\sqrt{3}+\tan A}{1-\sqrt{3} \tan A}-\frac{\sqrt{3}-\tan A}{1+\sqrt{3} \tan A} \\
& =\tan \mathrm{A}\\
& +\frac{\sqrt{3}+3 \tan A+\tan A+\sqrt{3} \tan ^2 A-\sqrt{3}+\tan A+3 \tan A-\sqrt{3} \tan ^2 A}{(1-\sqrt{3} \tan A)(1+\sqrt{3} \tan A)} \\
& =\tan A+\frac{8 \tan A}{1-3 \tan ^2 A} \\
& =\frac{\tan A-3 \tan ^3 \mathrm{~A}+8 \tan \mathrm{A}}{1-3 \tan ^2 \mathrm{~A}} \\
& =\frac{9 \tan A-3 \tan ^3 \mathrm{~A}}{1-3 \tan ^2 \mathrm{~A}} \\
& =3\left(\frac{3 \tan A-\tan ^3 \mathrm{~A}}{1-3 \tan ^2 \mathrm{~A}}\right) \\
& =3 \tan 3 \mathrm{~A} \\
& =\text { R.H.S. } \\
\end{aligned}$
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