Question

$\tan A+\tan (60°+A)-\tan (60°-A)$ is equal to

• A. $3\tan 3A$
• B. $\tan 3A$
• C. $cot3A$
• D. $\sin 3A$

Answer 1

The correct option is A

$3\tan 3A$

$\tan (60°+A)=\frac{\tan 60°+\tan A}{1-\tan 60°\tan A}$

$\tan (60°+A)=\frac{\sqrt{3}+\tan A}{1-\sqrt{3}\tan A}$ $\to \left(1\right)$

$\tan (60°-A)=\frac{\tan 60°-\tan A}{1+\tan 60°\tan A}$

$\tan (60°-A)=\frac{\sqrt{3}-\tan A}{1+\sqrt{3}\tan A}$ $\to \left(2\right)$

By subtracting equations $\left(1\right)$ and $\left(2\right)$, we get

$\tan (60°+A)-\tan (60°-A)=\frac{\sqrt{3}+\tan A}{1-\sqrt{3}\tan A}-\frac{\sqrt{3}-\tan A}{1+\sqrt{3}\tan A}$

$=\frac{\left(\sqrt{3}+\tan A\right)\left(1+\sqrt{3}\tan A\right)-\left(\sqrt{3}-\tan A\right)\left(1-\sqrt{3}\tan A\right)}{{\left(1\right)}^2-{\left(\sqrt{3}\tan A\right)}^2}$

$=\frac{\left(\sqrt{3}+\tan A\right)\left(1+\sqrt{3}\tan A\right)-\left(\sqrt{3}-\tan A\right)\left(1-\sqrt{3}\tan A\right)}{{\left(1\right)}^2-{\left(\sqrt{3}\tan A\right)}^2}$

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

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

$=\frac{8\tan A}{1-3{\tan }^{2}A}$

By adding $\tan A$ on both sides, we get

$\tan A+\tan \left(60°+A\right)-\tan \left(60°-A\right)=\tan A+\frac{8\tan A}{1-3{\tan }^{2}A}$

$\tan A+\tan \left(60°+A\right)-\tan \left(60°-A\right)=\frac{\tan A-3{\tan }^{3}A+8\tan A}{1-3{\tan }^{2}A}$

$\tan A+\tan \left(60°+A\right)-\tan \left(60°-A\right)=\frac{9\tan A-3{\tan }^{3}A}{1-3{\tan }^{2}A}$

$\tan A+\tan \left(60°+A\right)-\tan \left(60°-A\right)=\frac{3\left(3\tan A-{\tan }^{3}A\right)}{1-3{\tan }^{2}A}$

$\tan A+\tan \left(60°+A\right)-\tan \left(60°-A\right)=3\tan 3A$