If x cosec2 30° sec2 45°8 cos2 45° sin2 60°=tan2 60°−tan2 30°, then x =

Question:

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

Solution:

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 $(a)$