Question
If $A = 30^\circ,$ verify that $\sin 2 A=\frac{2 \tan A}{1+\tan ^2 A}$

Answer

$LHS =\sin 2A$
Putting $A = 30^\circ$ in $LHS$ and $RHS.,$ we get
LHS = sin 2 x 30° = sin 60° $=\frac{\sqrt{3}}{2}$
$\text { RHS }=\frac{2 \times \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}=\frac{2 \times \frac{1}{\sqrt{3}}}{1+\left(\frac{1}{\sqrt{3}}\right)^2}$
$=\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}} \cdot \frac{2}{\frac{4}{3}}$
$=\frac{2 \times 3}{\sqrt{3} \times 4}=\frac{\sqrt{3}}{4}$
Hence,
$LHS = RHS$
Hence proved.

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