Evaluate each of the following:$ 4\left(\sin ^{4} 60^{\circ}+\cos ^{4} 30^{\circ}\right)-3\left(\tan ^{2} 60^{\circ}-\tan ^{2} 45^{\circ}\right)+5 \cos ^{2} 45^{\circ} $

Given:

\( 4\left(\sin ^{4} 60^{\circ}+\cos ^{4} 30^{\circ}\right)-3\left(\tan ^{2} 60^{\circ}-\tan ^{2} 45^{\circ}\right)+5 \cos ^{2} 45^{\circ} \)

To do:

We have to evaluate \( 4\left(\sin ^{4} 60^{\circ}+\cos ^{4} 30^{\circ}\right)-3\left(\tan ^{2} 60^{\circ}-\tan ^{2} 45^{\circ}\right)+5 \cos ^{2} 45^{\circ} \).

Solution:  

We know that,

$\sin 60^{\circ}=\frac{\sqrt3}{2}$

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

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

$\tan 45^{\circ}=1$

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

$4\left(\sin ^{4} 60^{\circ}+\cos ^{4} 30^{\circ}\right)-3\left(\tan ^{2} 60^{\circ}-\tan ^{2} 45^{\circ}\right)+5 \cos ^{2} 45^{\circ}=4\left[\left(\frac{\sqrt{3}}{2}\right)^{4} +\left(\frac{\sqrt{3}}{2}\right)^{4}\right] -3\left[\left(\sqrt{3}\right)^{2} -( 1)^{2}\right] +5\left(\frac{1}{\sqrt{2}}\right)^{2}$

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

$=4\left(\frac{9}{16} +\frac{9}{16}\right) -3( 2) +\frac{5}{2}$

$=4\left(\frac{9+9}{16}\right) -6+\frac{5}{2}$

$=\frac{18}{4} -6+\frac{5}{2}$

$=\frac{18-6( 4) +5( 2)}{4}$

$=\frac{18-24+10}{4}$

$=\frac{4}{4}$

$=1$

Hence, $4\left(\sin ^{4} 60^{\circ}+\cos ^{4} 30^{\circ}\right)-3\left(\tan ^{2} 60^{\circ}-\tan ^{2} 45^{\circ}\right)+5 \cos ^{2} 45^{\circ}=1$. 

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Updated on: 10-Oct-2022

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