Question

The value of $32co{t}^2\frac{\pi }{4}-8se{c}^2\frac{\pi }{3}+8{cos}^3\frac{\pi }{6}$ is equal to:

• A. $3\sqrt{3}$
• B. $2\sqrt{3}$
• C. $\sqrt{3}$
• D. $6\sqrt{3}$

Answer 1

The correct option is A $3\sqrt{3}$

$\to 32{cot}^2\left(\pi /4\right)-8{sec}^2\left(\pi /3\right)+8{cos}^3\left(\pi /6\right)$

$\to 32{\left(1\right)}^2-8{\left(2\right)}^2+8{\left(\frac{\sqrt{3}}{2}\right)}^3=32-8\left(4\right)+3\sqrt{3}$

$\to 3\sqrt{3}$