$\left(\cos ^{2} 30^{\circ} \cos ^{2} 45^{\circ}+4 \sec ^{2} 60^{\circ}+\frac{1}{2} \cos ^{2} 90^{\circ}-2 \tan ^{2} 60^{\circ}\right)=?$
(a) $\frac{81}{8}$
(b) $\frac{83}{8}$
(C) $\frac{73}{8}$
(d) $\frac{75}{8}$
As we know that,
$\cos 30^{\circ}=\frac{\sqrt{3}}{2}$
$\cos 45^{\circ}=\frac{1}{\sqrt{2}}$
$\sec 60^{\circ}=2$
$\cos 90^{\circ}=0$
$\tan 60^{\circ}=\sqrt{3}$
By substituting these values, we get
$\left(\cos ^{2} 30^{\circ} \cos ^{2} 45^{\circ}+4 \sec ^{2} 60^{\circ}+\frac{1}{2} \cos ^{2} 90^{\circ}-2 \tan ^{2} 60^{\circ}\right)=\left(\frac{\sqrt{3}}{2}\right)^{2}\left(\frac{1}{\sqrt{2}}\right)^{2}+4(2)^{2}+\frac{1}{2}(0)^{2}-2(\sqrt{3})^{2}$
$=\left(\frac{3}{4}\right)\left(\frac{1}{2}\right)+4(4)-2(3)$
$=\frac{3}{8}+16-6$
$=\frac{3}{8}+10$
$=\frac{3+80}{8}$
$=\frac{83}{8}$
Hence, the correct option is (b).