Question

If $x\,sin \,\frac{\pi}{4}cos^{2}\, \frac{\pi}{3}=\frac{tan^{2}\left(\pi/3\right)cosec\left(\pi/6\right)}{sec\left(\pi/4\right)cot^{2}\left(\pi/6\right)}$, then $x=$

• A. $2$
• B. $4$
• C. $8$
• D. $16$

Answer 1

Answer: (C)

$x\,sin \,45^{\circ}\, cos^{2}60^{\circ}=\frac{tan^{2}\,60^{\circ}\,cosec\,30^{\circ}}{sec\,45^{\circ}\,cot^{2}\,30^{\circ}}$
$\Rightarrow x \cdot\frac{1}{\sqrt{2}}\cdot\frac{1}{4}=\frac{3\cdot 2}{\sqrt{2}\cdot 3}$
$\Rightarrow x=8$.