If $x\sin 45^\circ {\cos ^2}60^\circ = \frac{{{{\tan }^2}60^\circ {\rm{cosec}}30^\circ }}{{\sec 45^\circ {{\cot }^2}30^\circ }},$ then $x = $
If $x\sin 45^\circ {\cos ^2}60^\circ = \frac{{{{\tan }^2}60^\circ {\rm{cosec}}30^\circ }}{{\sec 45^\circ {{\cot }^2}30^\circ }},$ then $x = $
- A
$2$
- B
$4$
- C
$8$
- D
$16$
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- [IIT 1970]
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