Question:
If $x \sin 45^{\circ} \cos ^{2} 60^{\circ}=\frac{\tan ^{2} 60^{\circ} \operatorname{cosec} 30^{\circ}}{\sec 45^{\circ} \cot ^{2} 30^{\circ}}$, then $x=$
(a) 2
(b) 4
(c) 8
(d) 16
Solution:
(c) 8
We have:
$x \sin 45^{\circ} \cos ^{2} 60^{\circ}=\frac{\tan ^{2} 60^{\circ} \operatorname{cosec} 30^{\circ}}{\sec 45^{\circ} \cot ^{2} 30^{\circ}}$
$\Rightarrow x \times\left(\frac{1}{\sqrt{2}}\right) \times\left(\frac{1}{2}\right)^{2}=\frac{(\sqrt{3})^{2} \times(2)}{(\sqrt{2}) \times(\sqrt{3})^{2}}$
$\Rightarrow \frac{x}{4 \sqrt{2}}=\frac{6}{3 \sqrt{2}}$
$\Rightarrow x=\frac{6}{3 \sqrt{2}} \times 4 \sqrt{2}$
$\Rightarrow x=8$