If 3–√ tan θ=3 sin θ, find the value of sin2θ − cos2θ.

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

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

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

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

$\Rightarrow \cos \theta=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$

Therefore,

$\sin ^{2} \theta-\cos ^{2} \theta=1-\cos ^{2} \theta-\cos ^{2} \theta \quad\left(\right.$ since, $\left.\sin ^{2} \theta+\cos ^{2} \theta=1\right)$

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

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

$=\frac{1}{3}$

Hence, the value of the expression is $\frac{1}{3}$.