Question
Evaluate the following:
If A = 30°, verify that:
$\tan2\text{A}=\frac{2\tan\text{A}}{1-\tan^2\text{A}}$

Answer

$\text{A}=30^\circ$
$\Rightarrow2\text{A}=2\times30^\circ=60^\circ$
$\tan2\text{A}=\tan60^\circ=\sqrt{3}$
$\frac{2\tan\text{A}}{1-\tan^2\text{A}}=\frac{2\tan30^\circ}{1+\tan^230^\circ}$
$=\frac{2\times\Big(\frac{1}{\sqrt{3}}\Big)}{1-\Big(\frac{1}{\sqrt{3}}\Big)^2}=\frac{\Big(\frac{2}{\sqrt{3}}\Big)}{1-\frac13}$
$=\frac{\Big(\frac{2}{\sqrt{3}}\Big)}{\frac{2}{3}}=\Big(\frac{2}{\sqrt{3}}\Big)\times\frac32=\sqrt{3}$
$\therefore\ \tan2\text{A}=\frac{2\tan\text{A}}{1-\tan^2\text{A}}$

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