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

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

Vedclass · Accepted Answer

$8$

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(C) Given the equation: $x \sin 45^\circ \cos^2 60^\circ = \frac{	an^2 60^\circ \csc 30^\circ}{\sec 45^\circ \cot^2 30^\circ}$Substituting 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 \left( \frac{1}{\sqrt{2}} ight) \left( \frac{1}{2} ight)^2 = \frac{(\sqrt{3})^2 	imes 2}{\sqrt{2} 	imes (\sqrt{3})^2}$$x \left( \frac{1}{\sqrt{2}} ight) \left( \frac{1}{4} ight) = \frac{3 	imes 2}{\sqrt{2} 	imes 3}$$\frac{x}{4\sqrt{2}} = \frac{6}{3\sqrt{2}}$$\frac{x}{4\sqrt{2}} = \frac{2}{\sqrt{2}}$$x = \frac{2 	imes 4\sqrt{2}}{\sqrt{2}}$$x = 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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