Q.If \(x \sin 60^\circ \cos^2 30^\circ = \frac{\tan^2 45^\circ \sec 60^\circ}{\csc 60^\circ}\), find the value of \(x\).

Given: \[ x \sin 60^\circ \cos^2 30^\circ = \frac{\tan^2 45^\circ \sec 60^\circ}{\csc 60^\circ} \] Substituting the trigonometric values: \[ x \cdot \frac{\sqrt{3}}{2} \cdot \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1^2 \cdot 2}{\frac{2}{\sqrt{3}}} \] Simplifying: \[ x \cdot \frac{\sqrt{3}}{2} \cdot \frac{3}{4} = 2 \cdot \frac{\sqrt{3}}{2} \Rightarrow \frac{3\sqrt{3}x}{8} = \sqrt{3} \] Solving for \(x\): \[ x = \sqrt{3} \cdot \frac{8}{3\sqrt{3}} = \frac{8}{3} = 2\frac{2}{3} \] (Answer)
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