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Verify that $\sin 2A = \frac{2 \tan A}{1 + \tan^2 A}$, for $A = 30^\circ$.
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$LHS = \sin 60^\circ = \frac{\sqrt{3}}{2}$ ($\frac{1}{2}$ mark)
$RHS = \frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}} = \frac{\sqrt{3}}{2} = LHS$ ($1 + \frac{1}{2}$ marks)
$RHS = \frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}} = \frac{\sqrt{3}}{2} = LHS$ ($1 + \frac{1}{2}$ marks)