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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
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Solution
8
We have:
\[x \sin 45^\circ\cos^2 60^\circ = \frac{\tan^2 60^\circ cosec 30^\circ}{\sec45^\circ \cot^2 30^\circ}\]
\[ \Rightarrow x \times \left( \frac{1}{\sqrt{2}} \right) \times \left( \frac{1}{2} \right)^2 = \frac{\left( \sqrt{3} \right)^2 \times \left( 2 \right)}{\left( \sqrt{2} \right) \times \left( \sqrt{3} \right)^2}\]
\[ \Rightarrow \frac{x}{4\sqrt{2}} = \frac{6}{3\sqrt{2}}\]
\[ \Rightarrow x = \frac{6}{3\sqrt{2}} \times 4\sqrt{2}\]
\[ \Rightarrow x = 8\]
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