If $ \sqrt{3} \tan \theta=1 $, then find the value of $ \sin ^{2} \theta-\cos ^{2} \theta $.

Given:

\( \sqrt{3} \tan \theta=1 \)

To do:

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

Solution:  

$\sqrt{3} \tan \theta=1$       

$\Rightarrow \tan \theta=\frac{1}{\sqrt3}$

$\Rightarrow \tan \theta=\tan 30^{\circ}$

$\Rightarrow \theta=30^{\circ}$

Therefore,

$\sin ^{2} \theta-\cos ^{2} \theta=\sin ^{2} 30^{\circ}-\cos ^{2} 30^{\circ}$

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

$=\frac{1}{4}-\frac{3}{4}$

$=\frac{1-3}{4}$

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

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

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