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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.