Question

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

• A. $\cos {60}^{\circ }$
• B. $\sin {60}^{\circ }$
• C. $\tan {60}^{\circ }$
• D. $\sin {30}^{\circ }$

Answer 1

Answer: (C)

$\tan {60}^{\circ }$

Given,

$\frac{2\tan \left({30}^{\circ }\right)}{1-{\tan }^2\left({30}^{\circ }\right)}$

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

$=\frac{2}{\frac{2}{\sqrt{3}}}$

$=\sqrt{3}$

$=\tan 60$