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

Given:

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

To do:

We have to evaluate \( \frac{2}{3}\left(\cos ^{4} 30^{\circ}-\sin ^{4} 45^{\circ}\right)-3\left(\sin ^{2} 60^{\circ}-\sec ^{2} 45^{\circ}\right)+\frac{1}{4} \cot ^{2} 30^{\circ} \).

Solution:  

We know that,

$\cos 30^{\circ}=\frac{\sqrt3}{2}$

$\sin 45^{\circ}=\frac{1}{\sqrt2}$

$\sin 60^{\circ}=\frac{\sqrt3}{2}$

$\sec 45^{\circ}=\sqrt2$

$\cot 30^{\circ}=\sqrt3$

Therefore,

$\frac{2}{3}\left(\cos ^{4} 30^{\circ}-\sin ^{4} 45^{\circ}\right)-3\left(\sin ^{2} 60^{\circ}-\sec ^{2} 45^{\circ}\right)+\frac{1}{4} \cot ^{2} 30^{\circ}=\frac{2}{3}\left[\left(\frac{\sqrt{3}}{2}\right)^{4} -\left(\frac{1}{\sqrt{2}}\right)^{4}\right] -3\left[\left(\frac{\sqrt{3}}{2}\right)^{2} -\left(\sqrt{2}\right)^{2}\right] +\frac{1}{4}\left(\sqrt{3}\right)^{2}$

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

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

$=\frac{2}{3}\left(\frac{9-1( 4)}{16}\right) -3\left(\frac{3-8}{4}\right) +\frac{3}{4}$

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

$=\left(\frac{5}{3( 8)}\right) +\frac{15}{4} +\frac{3}{4}$

$=\frac{5}{24} +\frac{15+3}{4}$

$=\frac{5}{24} +\frac{18}{4}$

$=\frac{5( 1) +18( 6)}{24}$

$=\frac{5+108}{24}$

$=\frac{113}{24}$

Hence, $\frac{2}{3}\left(\cos ^{4} 30^{\circ}-\sin ^{4} 45^{\circ}\right)-3\left(\sin ^{2} 60^{\circ}-\sec ^{2} 45^{\circ}\right)+\frac{1}{4} \cot ^{2} 30^{\circ}=\frac{113}{24}$.