Question
If $\theta=30^\circ,$ verify that.
$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$
$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$
✓
Answer
$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$
Put $\theta=30^\circ$
$\Rightarrow\tan2\times30^\circ=\frac{2\tan30^\circ}{1-\tan^230^\circ}$
$\Rightarrow\tan60^\circ=\frac{2\times\frac{1}{\sqrt{3}}}{1-\big(\frac{1}{\sqrt{3}}\big)^2}$
$\Rightarrow\sqrt{3}=\frac{2}{\sqrt{3}}\times\frac{1}{1-\frac{1}{3}}$
$\Rightarrow\sqrt{3}=\frac{2}{\sqrt{3}}\times\frac{3}{2}$
$\Rightarrow\sqrt{3}=\sqrt{3}$
$\Rightarrow\text{L.H.S}=\text{R.H.S}$
Put $\theta=30^\circ$
$\Rightarrow\tan2\times30^\circ=\frac{2\tan30^\circ}{1-\tan^230^\circ}$
$\Rightarrow\tan60^\circ=\frac{2\times\frac{1}{\sqrt{3}}}{1-\big(\frac{1}{\sqrt{3}}\big)^2}$
$\Rightarrow\sqrt{3}=\frac{2}{\sqrt{3}}\times\frac{1}{1-\frac{1}{3}}$
$\Rightarrow\sqrt{3}=\frac{2}{\sqrt{3}}\times\frac{3}{2}$
$\Rightarrow\sqrt{3}=\sqrt{3}$
$\Rightarrow\text{L.H.S}=\text{R.H.S}$
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