Question

If A = ${60}^{\circ }$ verify the following .
4 . $sin2A=\frac{2tanA}{1+{tan}^{2}A}$

Answer 1

$\sin 2A=\frac{2\tan A}{1+{\tan }^{2}A}$

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

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

$=\sin {60}^{\circ }$

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

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

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

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

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

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

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

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

Hence verified