Question

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

Answer 1

Find the value of $\tan$$60°$:

Put $A=30°$

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

Given, $\tan 30°=\frac{1}{\sqrt{3}}$

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

Hence, the value of $\tan 60°$ is $\sqrt{3}$.