Question

Using the formula, $\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}$, find the value of $\tan {60}^{\circ }$, it being that $\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$.

Answer 1

Put $A={30}^{\circ }\Rightarrow 2A={60}^{\circ }$

$\tan {60}^{\circ }=\frac{2\tan {30}^{\circ }}{1-{\tan }^2{30}^{\circ }}$

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

$=\frac{2\times \frac{1}{\sqrt{3}}}{1-\frac{1}{3}}$

$=\sqrt{3}$

The value of $\tan {60}^{\circ }$ is $\sqrt{3}$.