Evaluate each of the following:$ \cot ^{2} 30^{\circ}-2 \cos ^{2} 60^{\circ}-\frac{3}{4} \sec ^{2} 45^{\circ}-4 \sec ^{2} 30^{\circ} $

Given:

\( \cot ^{2} 30^{\circ}-2 \cos ^{2} 60^{\circ}-\frac{3}{4} \sec ^{2} 45^{\circ}-4 \sec ^{2} 30^{\circ} \)

To do:

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

Solution:  

We know that,

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

$\cos 60^{\circ}=\frac{1}{2}$

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

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

Therefore,

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

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

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

$=\frac{3( 6) -1( 3) -3( 3) -16( 2)}{6}$

$=\frac{18-3-9-32}{6}$

$=\frac{-26}{6}$

$=\frac{-13}{3}$

Hence, $\cot ^{2} 30^{\circ}-2 \cos ^{2} 60^{\circ}-\frac{3}{4} \sec ^{2} 45^{\circ}-4 \sec ^{2} 30^{\circ}=\frac{-13}{3}$.