Evaluate each of the following

Solution:

We have,

$\sin ^{2} 30^{\circ} \cos ^{2} 45^{\circ}+4 \tan ^{2} 30^{\circ}+\frac{1}{2} \sin ^{2} 90^{\circ}-2 \cos ^{2} 90^{\circ}+\frac{1}{24} \cos ^{2} 0^{\circ}$.....(1)

Now,

$\sin 30^{\circ}=\frac{1}{2}, \cos 45^{\circ}=\frac{1}{\sqrt{2}}, \tan 30^{\circ}=\frac{1}{\sqrt{3}}, \sin 90^{\circ}=\cos 0^{\circ}=1, \cos 90^{\circ}=0$

So by substituting above values in equation (1)

We get,

$\sin ^{2} 30^{\circ} \cos ^{2} 45^{\circ}+4 \tan ^{2} 30^{\circ}+\frac{1}{2} \sin ^{2} 90^{\circ}-2 \cos ^{2} 90^{\circ}+\frac{1}{24} \cos ^{2} 0^{\circ}$

$=\left(\frac{1}{2}\right)^{2} \times\left(\frac{1}{\sqrt{2}}\right)^{2}+4 \times\left(\frac{1}{\sqrt{3}}\right)^{2}+\frac{1}{2} \times(1)^{2}-2 \times(0)^{2}+\frac{1}{24} \times(1)^{2}$

$=\frac{1}{4} \times \frac{1}{2}+4 \times \frac{1}{3}+\frac{1}{2} \times 1-2 \times 0+\frac{1}{24} \times 1$

$=\frac{1}{8}+\frac{4}{3}+\frac{1}{2}-0+\frac{1}{24}$

$=\frac{1}{8}+\frac{4}{3}+\frac{1}{2}+\frac{1}{24}$

LCM of 8, 3, 2 and 24 is 48

Therefore by taking LCM

We get,

$\sin ^{2} 30^{-} \cos ^{2} 45^{\circ}+4 \tan ^{2} 30^{\circ}+\frac{1}{2} \sin ^{2} 90^{\circ}-2 \cos ^{2} 90^{\circ}+\frac{1}{24} \cos ^{2} 0$

$=\frac{1 \times 6}{8 \times 6}+\frac{4 \times 16}{3 \times 16}+\frac{1 \times 24}{2 \times 24}+\frac{1 \times 2}{24 \times 2}$

$=\frac{6}{48}+\frac{64}{48}+\frac{24}{48}+\frac{2}{48}$

$=\frac{96}{48}$

In the above equation the first term $\frac{96}{48}$ gets reduced to 2

Therefore,

$\sin ^{2} 30^{\circ} \cos ^{2} 45^{\circ}+4 \tan ^{2} 30^{\circ}+\frac{1}{2} \sin ^{2} 90^{\circ}-2 \cos ^{2} 90^{\circ}+\frac{1}{24} \cos ^{2} 0^{\circ}=2$