Question

Verify, $\sin {60}^0=$ $\frac{2\tan 30}{1+{\tan }^{2}30}.$

Answer 1

L.H.S=$\sin {60}^0=\frac{\sqrt{3}}{2}$

R.H.S=$\frac{2\tan {30}^0}{1+{\tan }^2{30}^0}$

$=\frac{2\frac{1}{\sqrt{3}}}{1+\frac{1}{(\sqrt{3}{)}^2}}$

$=\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}$

$=\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}$

$=\frac{3}{2\sqrt{3}}$

$\frac{2\tan {30}^0}{1+{\tan }^2{30}^0}$$=\frac{\sqrt{3}}{2}$

Therefore, LHS=RHS

Hence, proved.