Evaluate each of the following

Solution:

We have,

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

Now,

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

So by substituting above values in equation (1)

We get,

$\frac{4}{\cot ^{2} 30^{\circ}}+\frac{1}{\sin ^{2} 60^{\circ}}-\cos ^{2} 45^{\circ}$

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

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

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

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

Now LCM of denominator of above expression is 6

Therefore by taking LCM we get,

$\frac{4}{\cot ^{2} 30^{\circ}}+\frac{1}{\sin ^{2} 60^{\circ}}-\cos ^{2} 45^{\circ}$

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

$=\frac{8}{6}+\frac{8}{6}-\frac{3}{6}$

$=\frac{8+8-3}{6}$

$=\frac{16-3}{6}$

$=\frac{13}{6}$

Hence,

$\frac{4}{\cot ^{2} 30^{\circ}}+\frac{1}{\sin ^{2} 60^{\circ}}-\cos ^{2} 45^{\circ}=\frac{13}{6}$