# Evaluate each of the following:$\frac{\tan ^{2} 60^{\circ}+4 \cos ^{2} 45^{\circ}+3 \sec ^{2} 30^{\circ}+5 \cos ^{2} 90^{\circ}}{\operatorname{cosec} 30^{\circ}+\sec 60^{\circ}-\cot ^{2} 30^{\circ}}$

Given:

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

To do:

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

Solution:

We know that,

$tan 60^{\circ}=\sqrt3$

$\cos 45^{\circ}=\frac{1}{\sqrt2}$

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

$\cos 90^{\circ}=0$

$cosec 30^{\circ}=2$

$\sec 60^{\circ}=2$

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

Therefore,

$\frac{\tan ^{2} 60^{\circ}+4 \cos ^{2} 45^{\circ}+3 \sec ^{2} 30^{\circ}+5 \cos ^{2} 90^{\circ}}{\operatorname{cosec} 30^{\circ}+\sec 60^{\circ}-\cot ^{2} 30^{\circ}}=\frac{\left(\sqrt{3}\right)^{2} +4\left(\frac{1}{\sqrt{2}}\right)^{2} +3\left(\frac{2}{\sqrt{3}}\right)^{2} +5( 0)^{2}}{( 2) +( 2) -\left(\sqrt{3}\right)^{2}}$

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

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

$=9$

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

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Updated on: 10-Oct-2022

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