Solution:

I) L.H.S = sin2A

= $sin2(30^{o})$

= $2sin60^{o}$

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

R.H.S = 2Sin2CosA

$= 2Sin30^{0}Cos30^{o}$

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

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

hence, SIn2A = 2SinACosA

ii) L.H.S = $cos2A = cos2(30^{o})$

$= cos60^{o}$

$= \frac{1}{2}$

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

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

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

$= \frac{1}{2}$

hence, $Cos2A = …