# Prove that

Question:

$\sin ^{2} 30^{\circ} \cos ^{2} 45^{\circ}+4 \tan ^{2} 30^{\circ}+\frac{1}{2} \sin ^{2} 90^{\circ}+\frac{1}{8} \cot ^{2} 60^{\circ}$

Solution:

As we know that,

$\sin 30^{\circ}=\frac{1}{2}$

$\cos 45^{\circ}=\frac{1}{\sqrt{2}}$

$\tan 30^{\circ}=\frac{1}{\sqrt{3}}$

$\sin 90^{\circ}=1$

$\cot 60^{\circ}=\frac{1}{\sqrt{3}}$

By substituting these values, we get

$\sin ^{2} 30^{\circ} \cos ^{2} 45^{\circ}+4 \tan ^{2} 30^{\circ}+\frac{1}{2} \sin ^{2} 90^{\circ}+\frac{1}{8} \cot ^{2} 60^{\circ}=\left(\frac{1}{2}\right)^{2} \times\left(\frac{1}{\sqrt{2}}\right)^{2}+4\left(\frac{1}{\sqrt{3}}\right)^{2}+\frac{1}{2}(1)^{2}+\frac{1}{8}\left(\frac{1}{\sqrt{3}}\right)^{2}$

$=\left(\frac{1}{4}\right) \times\left(\frac{1}{2}\right)+4\left(\frac{1}{3}\right)+\frac{1}{2}(1)+\frac{1}{8}\left(\frac{1}{3}\right)$

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

$=\frac{3+32+12+1}{24}$

$=\frac{48}{24}$

$=2$

Hence, $\sin ^{2} 30^{\circ} \cos ^{2} 45^{\circ}+4 \tan ^{2} 30^{\circ}+\frac{1}{2} \sin ^{2} 90^{\circ}+\frac{1}{8} \cot ^{2} 60^{\circ}=2$.