Evaluate each of the following

Solution:

We have,

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

Now,

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

So by substituting above values in equation (1)

We get,

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

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

$=3-2 \times \frac{1^{2}}{2^{2}}-\frac{3}{4} \times 2-4 \times \frac{2^{2}}{(\sqrt{3})^{2}}$

Now, in the third term 4 gets cancelled by 2 and 2 remains

Therefore,

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

$=3-2 \times \frac{1}{4}-\frac{3}{2}-4 \times \frac{4}{3}$

Now in the second term, 4 gets cancelled by 2 and 2 remains

Therefore,

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

$=3-\frac{1}{2}-\frac{3}{2}-4 \times \frac{4}{3}$

$=3-\frac{1}{2}-\frac{3}{2}-\frac{16}{3}$

Now, LCM of denominator in the above expression is 6

Therefore by taking LCM

We get,

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

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

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

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

$=\frac{18-12-32}{6}$

$=\frac{18-44}{6}$

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

Now in the above expression, $\frac{-26}{6}$ gets reduced to $\frac{-13}{3}$

Therefore,

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