Question

If $\dfrac{2\tan30^\circ}{1+\tan^2 30^\circ} = \dfrac{2\tan30^\circ}{\sqrt{1 - \tan^2 30^\circ}}$, then $x : y =$

• A. 1 : 1
• B. 1 : 2
• C. 2 : 1
• D. 4 : 1

Hint:

Know your trigonometric identities and values for standard angles.

Answer 1

The Correct Option is C

$\tan 30^\circ = \dfrac{1}{\sqrt{3}}$
LHS = $\dfrac{2 \cdot \dfrac{1}{\sqrt{3}}}{1 + \dfrac{1}{3}} = \dfrac{2/\sqrt{3}}{4/3} = \dfrac{6}{4\sqrt{3}}$
RHS = $\dfrac{2 \cdot \dfrac{1}{\sqrt{3}}}{\sqrt{1 - \dfrac{1}{3}}} = \dfrac{2/\sqrt{3}}{\sqrt{2/3}} = \dfrac{2}{\sqrt{3}} \cdot \sqrt{3/2} = \sqrt{2}$
Matching LHS and RHS gives $x : y = 2 : 1$