120
Verify that $\cos 2A = \frac{1 - \tan^2 A}{1 + \tan^2 A}$ for $A = 30^\circ$.
Show SolutionHide Solution↓
$LHS = \cos 60^\circ = \frac{1}{2}$
$RHS = \frac{1 - \tan^2 30^\circ}{1 + \tan^2 30^\circ} = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{1}{2} = LHS$
$RHS = \frac{1 - \tan^2 30^\circ}{1 + \tan^2 30^\circ} = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{1}{2} = LHS$