If $ \theta=30^{\circ} $, verify that:$ \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta} $

Given:

\( \theta=30^{\circ} \)

To do:

We have to verify that \( \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta} \).

Solution:  

\( \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta} \)

This implies,

\( \tan 2(30^{\circ})=\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}} \)

\( \tan 60^{\circ}=\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}} \)

We know that,

$\tan 60^{\circ}=\sqrt3$

$\tan 30^{\circ}=\frac{1}{\sqrt3}$

Let us consider LHS,

$\tan 2 \theta=\tan 60^{\circ}$

$=\sqrt3$

Let us consider RHS,

$\frac{2 \tan \theta}{1-\tan ^{2} \theta}=\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}$

$=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}$

$=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}$

$=\frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}}$

$=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$

$=\frac{2}{\sqrt{3}} \times \frac{3}{2}$

$=\frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3}}$

$=\sqrt{3}$

LHS $=$ RHS

Hence proved.

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Updated on: 10-Oct-2022

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