Question

If x sin 45° cos² 60° = \[\frac{\tan^2 60^\circ cosec30^\circ}{\sec45^\circ \cot^{2^\circ} 30^\circ}\], then x =

 

Options

• 2

• 4

• 8

• 16

Answer 1

8
We have: 
\[x \sin 45^\circ\cos^2 60^\circ = \frac{\tan^2 60^\circ cosec 30^\circ}{\sec45^\circ \cot^2 30^\circ}\]

\[ \Rightarrow x \times \left( \frac{1}{\sqrt{2}} \right) \times \left( \frac{1}{2} \right)^2 = \frac{\left( \sqrt{3} \right)^2 \times \left( 2 \right)}{\left( \sqrt{2} \right) \times \left( \sqrt{3} \right)^2}\]

\[ \Rightarrow \frac{x}{4\sqrt{2}} = \frac{6}{3\sqrt{2}}\]

\[ \Rightarrow x = \frac{6}{3\sqrt{2}} \times 4\sqrt{2}\]

\[ \Rightarrow x = 8\]