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

A = 30^o
⇒ 2A = 2 $\times$ 30^o = 60^o

By substituting the value of the given T-ratio, we get:
$$\begin{gathered} \tan 2A=\frac{2\tan A}{1-{\tan }^{2}A} \\ \Rightarrow \tan {60}^o=\frac{2\tan {30}^o}{1-{\tan }^2{30}^o}=\frac{2\times {\displaystyle \frac{1}{\sqrt{3}}}}{1-{\left({\displaystyle \frac{1}{\sqrt{3}}}\right)}^2}=\frac{{\displaystyle \frac{2}{\sqrt{3}}}}{1-{\displaystyle \frac{1}{3}}}=\frac{{\displaystyle \frac{2}{\sqrt{3}}}}{{\displaystyle \frac{2}{3}}}=\frac{2}{\sqrt{3}}\times \frac{3}{2}=\sqrt{3} \end{gathered}$$
∴ tan 60^o = $\sqrt{3}$