Question

Verify that :
$\sin {60}^{\circ }=\frac{2\tan {30}^{\circ }}{1+{\tan }^2{30}^{\circ }}=\frac{\sqrt{3}}{2}$

Answer 1

RHS = $\frac{2\tan {30}^{\circ }}{1+{\tan }^2{30}^{\circ }}=\frac{2\times \frac{1}{\sqrt{3}}}{1+{\left(\frac{1}{\sqrt{3}}\right)}^2}=\frac{2/\sqrt{3}}{1+\frac{1}{3}}=\frac{2/\sqrt{3}}{4/3}=\frac{2}{\sqrt{3}}\times \frac{3}{4}=\frac{\sqrt{3}}{2}$
$\therefore LHS=RHS\sin {60}^{\circ }=\frac{2\tan {30}^{\circ }}{1+{\tan }^{2}30}=\frac{\sqrt{3}}{2}$