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

Given:

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

To do:

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

Solution:  

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

This implies,

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

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

We know that,

$\cos 60^{\circ}=\frac{1}{2}$

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

Let us consider LHS,

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

$=\frac{1}{2}$

Let us consider RHS,

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

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

$=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}$

$=\frac{\frac{3-1}{3}}{\frac{3+1}{3}}$

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

$=\frac{2}{4}$

$=\frac{1}{2}$

LHS $=$ RHS

Hence proved. 

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Updated on: 10-Oct-2022

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