Question

If A = ${60}^{\circ }$ verify the following .
$tan2A=\frac{2tanA}{1-{tan}^{2}A}$

Answer 1

$\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}$

L.H.S$=\tan (2\times {60}^{\circ })$

$=\tan ({180}^{\circ }-{60}^{\circ })$

$=-\tan {60}^{\circ }$

$=-\sqrt{3}$ ......$\left(1\right)$

R.H.S$=\frac{2\tan A}{1-{\tan }^{2}A}$

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

$=\frac{2\sqrt{3}}{1-{\left(\sqrt{3}\right)}^2}$

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

$=\frac{2\sqrt{3}}{-2}=-\sqrt{3}$ ......$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

L.H.S$=$R.H.S

Hence verified