Question

please solve the question no. 10

Answer 1

took A = 30°
3/√3 = √3

(1/√3)² = 1/3

Answer 2

Given,

$\tan2A=\frac{2\tan A}{1-\tan^{2}A}\\\:\\\text{Putting  A=30}\\\;\\\\tan2*30=\frac{2\tan30}{1-\tan^{2}30}\\\;\\\tan60=2\frac{\frac{1}{\sqrt{3}}}{1-(\frac{1}{\sqrt{3}})^{2}}\\\;\\\tan60=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}\\\;\\\tan60=\frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}}\\\;\\\tan60=\frac{2*3}{2*\sqrt{3}}\\\;\\\tan60=\sqrt{3}$