If x sin 45^circ cos^2 60^circ = frac{tan^2 6 | vedclass

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(C) Given the equation: $x \sin 45^\circ \cos^2 60^\circ = \frac{	an^2 60^\circ \csc 30^\circ}{\sec 45^\circ \cot^2 30^\circ}$Substitute the trigonom

Vedclass · Accepted Answer

$8$

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(C) Given the equation: $x \sin 45^\circ \cos^2 60^\circ = \frac{	an^2 60^\circ \csc 30^\circ}{\sec 45^\circ \cot^2 30^\circ}$Substitute the trigonometric values: $\sin 45^\circ = \frac{1}{\sqrt{2}}$,$\cos 60^\circ = \frac{1}{2}$,$	an 60^\circ = \sqrt{3}$,$\csc 30^\circ = 2$,$\sec 45^\circ = \sqrt{2}$,$\cot 30^\circ = \sqrt{3}$.$x \cdot \frac{1}{\sqrt{2}} \cdot (\frac{1}{2})^2 = \frac{(\sqrt{3})^2 \cdot 2}{\sqrt{2} \cdot (\sqrt{3})^2}$$x \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{4} = \frac{3 \cdot 2}{\sqrt{2} \cdot 3}$$\frac{x}{4\sqrt{2}} = \frac{6}{3\sqrt{2}}$$\frac{x}{4\sqrt{2}} = \frac{2}{\sqrt{2}}$$x = 2 \cdot 4 = 8$.

If $x \sin 45^\circ \cos^2 60^\circ = \frac{\tan^2 60^\circ \csc 30^\circ}{\sec 45^\circ \cot^2 30^\circ}$,then $x = $

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