If $ \sqrt{3} \tan \theta=3 \sin \theta $, find the value of $ \sin ^{2} \theta-\cos ^{2} \theta $


Given:

\( \sqrt{3} \tan \theta=3 \sin \theta \)

To do:

We have to find the value of \( \sin ^{2} \theta-\cos ^{2} \theta \).

Solution:  

We know that,

$\tan \theta=\frac{\sin \theta}{\cos \theta}$

This implies,

$\sqrt{3} \tan \theta=3 \sin \theta$

$\sqrt{3} \frac{\sin \theta}{\cos \theta}=3 \sin \theta$

$\sqrt{3}=3 \cos \theta$

$\cos \theta=\frac{\sqrt3}{3}$

$\cos \theta=\frac{1}{\sqrt3}$

$\Rightarrow \sin^{2} \theta=1-\cos^2 \theta$

$=1-(\frac{1}{\sqrt3})^2$

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

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

$=\frac{2}{3}$

Therefore,

$\sin ^{2} \theta-\cos ^{2} \theta=\frac{2}{3}-(\frac{1}{\sqrt3})^2$

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

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

$=\frac{1}{3}$

The value of \( \sin ^{2} \theta-\cos ^{2} \theta \) is $\frac{1}{3}$.  

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Updated on: 10-Oct-2022

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