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