$\dfrac{\text{2 tan 30°}}{1 + \text{tan}^2 \text{ 30°}}$ is equal to :

1. sin 60°

2. cos 60°

3. sec 60°

4. cosec 60°

$\dfrac{\text{2 tan 30°}}{1 + \text{tan}^2 \text{ 30°}} = \dfrac{2 \times \dfrac{1}{\sqrt{3}}}{1 + \Big(\dfrac{1}{\sqrt{3}}\Big)^2}\\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{1 + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{3 + 1}{3}}\\[1em] = \dfrac{2 \times 3}{4 \times {\sqrt{3}}}\\[1em] = \dfrac{6}{4{\sqrt{3}}}\\[1em] = \dfrac{3}{2{\sqrt{3}}}\\[1em] = \dfrac{{\sqrt{3}}}{2}\\[1em] = \text{sin} 60°$

Hence, option 1 is the correct option.