Question

Evaluate :
$\frac{2}{3}({\cos }^4{30}^{\circ }-{\sin }^4{45}^{\circ })-3({\sin }^2{60}^{\circ }-{\sec }^2{45}^{\circ })+\frac{1}{4}{\cot }^2{30}^{\circ }$

Answer 1

Given $\frac{2}{3}({\cos }^4{30}^{\circ }-{\sin }^4{45}^{\circ })-3({\sin }^2{60}^{\circ }-{\sec }^2{45}^{\circ })+\frac{1}{4}{\cot }^2{30}^{\circ }$

we know

$\cos {30}^{\circ }=\sin {60}^{\circ }=\frac{\sqrt{3}}{2},\sin {45}^{\circ }=\frac{1}{\sqrt{2}},\sec {45}^{\circ }=\sqrt{2},\cot {30}^{\circ }=\sqrt{3}$

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

$=\frac{2}{3}(\left(\frac{9}{16}\right)-\left(\frac{1}{4}\right))-3(\left(\frac{3}{4}\right)-\left(2\right))+\frac{1}{4}\left(3\right)$

$=\frac{2}{3}\left(\frac{5}{16}\right)-3\left(\frac{-5}{4}\right)+\frac{1}{4}\left(3\right)$

$=\frac{5}{24}+\frac{90}{24}+\frac{18}{24}$

$=\frac{113}{24}$