Question

The value of $\frac{2tan{30}^{\circ }}{1-ta{n}^2{30}^{\circ }}$ equals to

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

Answer 1

The correct option is C $tan$ ${60}^{\circ }$

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

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

$=\sqrt{3}$

$=\tan 60°$

Hence, the answer is $\tan 60°.$