If $\sqrt{5} \tan \theta=5 \sin \theta$, what is the value of $\left(\sin ^{2} \theta\right.$ $\left.-\cos ^{2} \theta\right) ?$

$\sqrt{5} \tan \theta=5 \sin \theta$

$\Rightarrow \sqrt{5} \frac{\sin \theta}{\cos \theta}=5 \sin \theta$

$\Rightarrow \frac{\sqrt{5} \sin \theta}{\cos \theta}-5 \sin \theta=0$

$\Rightarrow \sqrt{5} \sin \theta\left(\frac{1}{\cos \theta}-\sqrt{5}\right)=0$

$\Rightarrow \sqrt{5} \sin \theta\left(\frac{1}{\cos \theta}-\sqrt{5}\right)=0$

$\Rightarrow \cos \theta=\frac{1}{\sqrt{5}}$ or $\sin \theta=0$

$\therefore \sin \theta=\sqrt{1-\cos ^{2} \theta}=\sqrt{1-\frac{1}{5}}=\sqrt{\frac{4}{5}}=\frac{2}{\sqrt{5}}$

$\therefore \sin ^{2} \theta-\cos ^{2} \theta=\frac{4}{5}-\frac{1}{5}=\frac{3}{5}$