Question

Prove that:

$\tan {60}^{\circ }=\frac{2\tan {30}^{\circ }}{1-{\tan }^2{30}^{\circ }}$

Answer 1

To prove: $\tan {60}^o=\frac{2\tan {30}^o}{1-{\tan }^2{30}^o}$

Solving the R.H.S.:

$\frac{2tan{30}^o}{1-{\tan }^2{30}^o}$

We know that, $\tan {30}^{{}^o}=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{2\times \frac{1}{\sqrt{3}}}{1-(\frac{1}{\sqrt{3}}{)}^2}$

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

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

We know that $\tan {60}^o=\sqrt{3}$

$\therefore$R.H.S.$=\tan {60}^o$

Hence, L.H.S. $=$ R.H.S. $=\tan {60}^o$

$Henceproved$