If √3 tanθ = 1, then find the value of

Given that, $\sqrt{3} \tan \theta=1$

$\Rightarrow$ $\tan \theta=\frac{1}{\sqrt{3}}=\tan 30^{\circ}$

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

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

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

$=\frac{1}{4}-\frac{3}{4}=\frac{1-3}{4}=-\frac{2}{4}=-\frac{1}{2}$