Evaluate each of the following

We have,

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

Now,

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

So by substituting above values in equation (1)

We get,

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

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

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

$=4\left(\frac{9}{16}+\frac{9}{16}\right)-3(2)+5 \times \frac{1}{2}$

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

$=4\left(\frac{18}{16}\right)-6+\frac{5}{2}$

Now, $\frac{18}{16}$ gets reduced to $\frac{9}{8}$

Therefore,

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

$=4\left(\frac{9}{8}\right)-6+\frac{5}{2}$

$=\frac{36}{8}-6+\frac{5}{2}$

Now, $\frac{36}{8}$ gets reduced to $\frac{9}{2}$

Therefore,

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

$=\frac{9}{2}-6+\frac{5}{2}$

Now by taking LCM

We get,

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

$=\frac{9}{2}-\frac{6 \times 2}{1 \times 2}+\frac{5}{2}$

$=\frac{9}{2}-\frac{12}{2}+\frac{5}{2}$

$=\frac{9-12+5}{2}$

$=\frac{14-12}{2}$

$=\frac{2}{2}$

$=1$

Therefore,

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