Solve this

$\left(\frac{2 \tan ^{2} 30^{\circ} \sec ^{2} 52^{\circ} \sin ^{2} 38^{\circ}}{\operatorname{cosec}^{2} 70^{\circ}-\tan ^{2} 20^{\circ}}\right)$

$=\left(\frac{2 \tan ^{2} 30^{\circ}\left(\sec \left(90^{\circ}-38^{\circ}\right)\right)^{2} \sin ^{2} 38^{\circ}}{\left(\operatorname{cosec}\left(90^{\circ}-20^{\circ}\right)\right)^{2}-\tan ^{2} 20^{\circ}}\right)$

$=\left(\frac{2 \tan ^{2} 30^{\circ} \operatorname{cosec}^{2} 38^{\circ} \sin ^{2} 38^{\circ}}{\sec ^{2} 20^{\circ}-\tan ^{2} 20^{\circ}}\right) \quad\left(\because \sec \left(90^{\circ}-\theta\right)=\operatorname{cosec} \theta\right.$ and $\left.\operatorname{cosec}\left(90^{\circ}-\theta\right)=\sec \theta\right)$

$=\left(\frac{2 \tan ^{2} 30^{\circ} \operatorname{cosec}^{2} 38^{\circ} \sin ^{2} 38^{\circ}}{1}\right) \quad$ (using the identity: $\sec ^{2} \theta-\tan ^{2} \theta=1$ )

$=2 \tan ^{2} 30^{\circ} \frac{1}{\sin ^{2} 38^{\circ}} \sin ^{2} 38^{\circ} \quad\left(\because \operatorname{cosec} \theta=\frac{1}{\sin \theta}\right)$

$=2 \tan ^{2} 30^{\circ}$

$=2\left(\frac{1}{\sqrt{3}}\right)^{2} \quad\left(\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right)$

$=\frac{2}{3}$

Hence, the correct option is (c).