Solve this

Solution:

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).