# Evaluate each of the following

Question:

Evaluate each of the following

$\frac{\tan ^{2} 60^{\circ}+4 \cos ^{2} 45^{\circ}+3 \sec ^{2} 30^{\circ}+5 \cos ^{2} 90^{\circ}}{\operatorname{cosec} 30^{\circ}+\sec 60^{\circ}-\cot ^{2} 30^{\circ}}$

Solution:

We have,

$\frac{\tan ^{2} 60^{\circ}+4 \cos ^{2} 45^{\circ}+3 \sec ^{2} 30^{\circ}+5 \cos ^{2} 90^{\circ}}{\operatorname{cosec} 30^{\circ}+\sec 60^{\circ}-\cot ^{2} 30^{\circ}}$.....(1)

Now,

$\tan 60^{\circ}=\cot 30^{\circ}=\sqrt{3}, \cos 45^{\circ}=\frac{1}{\sqrt{2}}, \sec 30^{\circ}=\frac{2}{\sqrt{3}}, \sec 60^{\circ}=2, \operatorname{cosec} 30^{\circ}=2, \cos 90^{\circ}=0$

So by substituting above values in equation (1)

We get,

$\frac{\tan ^{2} 60^{\circ}+4 \cos ^{2} 45^{\circ}+3 \sec ^{2} 30^{\circ}+5 \cos ^{2} 90^{\circ}}{\operatorname{cosec} 30^{\circ}+\sec 60^{\circ}-\cot ^{2} 30^{\circ}}$

$=\frac{(\sqrt{3})^{2}+4\left(\frac{1}{\sqrt{2}}\right)^{2}+3\left(\frac{2}{\sqrt{3}}\right)^{2}+5(0)^{2}}{2+2-(\sqrt{3})^{2}}$

$=\frac{3+\frac{4}{2}+3\left(\frac{4}{3}\right)+0}{4-3}$

Now,

3 gets cancel in numerator and we get,

$\frac{\tan ^{2} 60^{\circ}+4 \cos ^{2} 45^{\circ}+3 \sec ^{2} 30^{\circ}+5 \cos ^{2} 90^{\circ}}{\operatorname{cosec} 30^{\circ}+\sec 60^{\circ}-\cot ^{2} 30^{\circ}}$

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

$=\frac{7+\frac{4}{2}}{1}$

Now, $\frac{4}{2}$ in the numerator get reduced to 2 and we get,

$\frac{\tan ^{2} 60^{\circ}+4 \cos ^{2} 45^{\circ}+3 \sec ^{2} 30^{\circ}+5 \cos ^{2} 90^{\circ}}{\operatorname{cosec} 30^{\circ}+\sec 60^{\circ}-\cot ^{2} 30^{\circ}}$

$=\frac{7+2}{1}$

$=\frac{9}{1}$

$=9$

$=9$

Therefore,

$\frac{\tan ^{2} 60^{\circ}+4 \cos ^{2} 45^{\circ}+3 \sec ^{2} 30^{\circ}+5 \cos ^{2} 90^{\circ}}{\operatorname{cosec} 30^{\circ}+\sec 60^{\circ}-\cot ^{2} 30^{\circ}}=9$