Using the formula, tan 2 A
Question:

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

Solution:

$A=30^{\circ}$

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

By substituting the value of the given T-ratio, we get:

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

$\Rightarrow \tan 60^{\circ}=\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=\frac{2 \times \frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}=\frac{2}{\sqrt{3}} \times \frac{3}{2}=\sqrt{3}$

$\therefore \tan 60^{\circ}=\sqrt{3}$