Prove that:
$2 \sin ^{2} \frac{\pi}{6}+\csc ^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}=\frac{3}{2}$

$L.H.S. = 2 \sin ^{2} \frac{\pi}{6} + \csc ^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}$
$= 2 \left( \frac{1}{2} \right)^{2} + \csc ^{2} \left( \pi + \frac{\pi}{6} \right) \left( \frac{1}{2} \right)^{2}$
$= 2 \times \frac{1}{4} + \left( -\csc \frac{\pi}{6} \right)^{2} \left( \frac{1}{4} \right)$
$= \frac{1}{2} + (-2)^{2} \left( \frac{1}{4} \right)$
$= \frac{1}{2} + \frac{4}{4} = \frac{1}{2} + 1 = \frac{3}{2}$
$= R.H.S.$