Question

Evaluate : $\frac{2\tan {30}^{\circ }}{1-{\tan }^2{30}^{\circ }}$

• A. $0$
• B. $1$
• C. $\sqrt{2}$
• D. $\sqrt{3}$

Answer 1

Answer: (D)

$\sqrt{3}$

We know that $\tan 2A$ $=\frac{2\tan A}{1-{\tan }^{2}A}$
Here $A={30}^0$
Substituting, we get
$\frac{2\tan {30}^0}{1-{\tan }^2{30}^0}$
$=\tan \left(2\right({30}^0\left)\right)$
$=\tan \left({60}^0\right)$
$=\sqrt{3}$