If A = 30°, then prove that :

$\text{sin 2 A} = \text{2 sin A cos A} = \dfrac{\text{2 tan A}}{1 + \text{tan}^2 \text{ A}}$

$\text{sin 2 A} = \text{2 sin A cos A} = \dfrac{\text{2 tan A}}{1 + \text{tan}^2 \text{ A}}$

$\text{1st term} = \text{sin 2 A} = \text{sin (2 x 30°)} = \text{sin 60°} = \dfrac{\sqrt3}{2}$

$\text{2nd term} = \text{2 sin A cos A}\\[1em] = 2 \times \text{sin 30°} \times \text{cos 30°}\\[1em] = 2 \times \dfrac{1}{2} \times \dfrac{\sqrt3}{2}\\[1em] = \cancel{2} \times \dfrac{1}{\cancel{2}} \times \dfrac{\sqrt3}{2}\\[1em] = \dfrac{\sqrt3}{2}$

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

∴ 1st term = 2nd term = 3rd term

Hence, $\text{sin 2 A} = \text{2 sin A cos A} = \dfrac{\text{2 tan A}}{1 + \text{tan}^2 \text{ A}}$.