Q.If \(x\sin45^\circ \cos45^\circ \tan60^\circ = \tan^2 45^\circ - \cos^2 60^\circ\), then what is the value of \(x\)? (a) 1 (b) \(\cfrac{2}{\sqrt3}\) (c) \(\cfrac{1}{\sqrt3}\) (d) \(\cfrac{\sqrt3}{2}\)
Answer: D
\(x\sin45^\circ \cos45^\circ \tan60^\circ = \tan^2 45^\circ - \cos^2 60^\circ\) i.e., \(x \times \cfrac{1}{\sqrt{2}} \times \cfrac{1}{\sqrt{2}} \times \sqrt{3} = 1^2 - \left(\cfrac{1}{2}\right)^2\) i.e., \(x \times \cfrac{\sqrt{3}}{2} = 1 - \cfrac{1}{4}\) i.e., \(x = \cfrac{3}{4} \times \cfrac{2}{\sqrt{3}} = \cfrac{\sqrt{3}}{2}\)
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