Question : The value of $x$ in the expression $\tan^{2}\frac{\pi }{4}-\cos^{2}\frac{\pi }{3}=x\sin\frac{\pi }{4}\cos\frac{\pi }{4}\tan\frac{\pi }{3}$ is:
Option 1: $\frac{2}{\sqrt{3}}$
Option 2: $\frac{3\sqrt{3}}{4}$
Option 3: $\frac{1}{\sqrt{3}}$
Option 4: $\frac{\sqrt{3}}{2}$
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Correct Answer: $\frac{\sqrt{3}}{2}$
Solution : $\tan^{2}\frac{\pi }{4}-\cos^{2}\frac{\pi }{3}=x\sin\frac{\pi }{4}\cos\frac{\pi }{4}\tan\frac{\pi }{3}$ Taking L.H.S. $⇒\tan^{2}\frac{\pi}{4} - \cos^{2}\frac{\pi}{3} = 1 - \left(\frac{1}{2}\right)^{2} = 1 - \frac{1}{4} = \frac{3}{4}$ Taking R.H.S. $⇒x\sin\frac{\pi}{4}\cos\frac{\pi}{4}\tan\frac{\pi}{3} = x \times \frac{1}{2} \times\sqrt{3} = \frac{x\sqrt{3}}{2}$ $⇒\frac{3}{4} = \frac{x\sqrt{3}}{2}$ $⇒x = \frac{3}{4} \times \frac{2}{\sqrt{3}} = \frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2}$ Hence, the correct answer is $\frac{\sqrt{3}}{2}$.
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