If $\frac{x \operatorname{cosec}^{2} 30^{\circ} \sec ^{2} 45^{\circ}}{8 \cos ^{2} 45^{\circ} \sin ^{2} 60^{\circ}}=\tan ^{2} 60^{\circ}-\tan ^{2} 30^{\circ}$, then $x=$
(a) 1
(b) $-1$
(c) 2
(d) 0
We have: $\frac{x \operatorname{cosec}^{2} 30^{\circ} \sec ^{2} 45^{\circ}}{8 \cos ^{2} 45^{\circ} \sin ^{2} 60^{\circ}}=\tan ^{2} 60^{\circ}-\tan ^{2} 30^{\circ}$
Here we have to find the value of $x$
As we know that $\left[\begin{array}{l}\cos 45^{\circ}=\frac{1}{\sqrt{2}} \\ \sec 45^{\circ}=\sqrt{2} \\ \tan 30^{\circ}=\frac{1}{\sqrt{3}} \\ \tan 60^{\circ}=\sqrt{3} \\ \cos 30^{\circ}=\frac{\sqrt{3}}{2} \\ \operatorname{cosec} 30^{\circ}=2\end{array}\right]$
So
$\Rightarrow \frac{x \operatorname{cosec}^{2} 30^{\circ} \sec ^{2} 45^{\circ}}{8 \cos ^{2} 45^{\circ} \sin ^{2} 60^{\circ}}=\tan ^{2} 60^{\circ}-\tan ^{2} 30^{\circ}$
$\Rightarrow \frac{x \times 4 \times 2}{8 \times \frac{1}{2} \times \frac{3}{4}}=3-\frac{1}{3}$
$\Rightarrow \frac{8 x}{3}=\frac{8}{3}$
$\Rightarrow x=1$
Hence the correct option is $(a)$