若 $\tan ^{2} x-\sin ^{2} x=\frac{16}{5}$, 则 $\tan ^{2} x \sin ^{2} x=$ \$ \qquad \$ .
$\tan ^{2} x \sin ^{2} x=\tan ^{2} x\left(1-\cos ^{2} x\right)=\tan ^{2} x-\tan ^{2} x \cos ^{2} x=\tan ^{2} x-\sin ^{2} x=\frac{16}{5}$.