Prove that :

$\text{sin }(2 \times 30°) = \dfrac{2 \text{ tan 30°}}{1 + \text{tan}^2 \text{ 30°}}$

$\text{sin }(2 \times 30°) = \dfrac{2 \text{ tan 30°}}{1 + \text{tan}^2 \text{ 30°}}$

L.H.S. = sin (2 x 30°) = sin 60° = $\dfrac{\sqrt3}{2}$

R.H.S.

$= \dfrac{2 \text{ tan 30°}}{1 + \text{tan}^2 \text{ 30°}}\\[1em] = \dfrac{2 \times \dfrac{1}{\sqrt3}}{1 + \Big(\dfrac{1}{\sqrt3}\Big)^2}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{1 + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{\dfrac{3 + 1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{\dfrac{4}{3}}\\[1em] = \dfrac{2 \times 3}{4 \times \sqrt3}\\[1em] = \dfrac{\sqrt3}{2}\\[1em]$

∴ L.H.S. = R.H.S.

Hence, $\text{sin }(2 \times 30°) = \dfrac{2 \text{ tan 30°}}{1 + \text{tan}^2 \text{ 30°}}$