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

Given:

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

To do:

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

Solution:  

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

This implies,

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

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

We know that,

$\sin 60^{\circ}=\frac{\sqrt3}{2}$

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

Let us consider LHS,

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

$=\frac{\sqrt3}{2}$

Let us consider RHS,

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

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

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

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

$=\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}$

$=\frac{2}{\sqrt{3}} \times \frac{3}{4}$

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

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

LHS $=$ RHS

Hence proved. 

Tutorialspoint

Updated on: 10-Oct-2022

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