Question

$\left({\cos }^{2}30°{\cos }^{2}45°+4{\sec }^{2}60°+\frac{1}{2}{\cos }^{2}90°-2{\tan }^{2}60°\right)=?$

(a) $\frac{81}{8}$

(b) $\frac{83}{8}$

(c) $\frac{73}{8}$

(d) $\frac{75}{8}$

Answer 1

As we know that,
$$\begin{gathered} \cos 30°=\frac{\sqrt{3}}{2} \\ \cos 45°=\frac{1}{\sqrt{2}} \\ \sec 60°=2 \\ \cos 90°=0 \\ \tan 60°=\sqrt{3} \\ \mathrm{By}\mathrm{substituting}\mathrm{these}\mathrm{values},\mathrm{we}\mathrm{get} \\ \left({\cos }^{2}30°{\cos }^{2}45°+4{\sec }^{2}60°+\frac{1}{2}{\cos }^{2}90°-2{\tan }^{2}60°\right)={\left(\frac{\sqrt{3}}{2}\right)}^2{\left(\frac{1}{\sqrt{2}}\right)}^2+4{\left(2\right)}^2+\frac{1}{2}{\left(0\right)}^2-2{\left(\sqrt{3}\right)}^2 \\ =\left(\frac{3}{4}\right)\left(\frac{1}{2}\right)+4\left(4\right)-2\left(3\right) \\ =\frac{3}{8}+16-6 \\ =\frac{3}{8}+10 \\ =\frac{3+80}{8} \\ =\frac{83}{8} \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{b}\right). \end{gathered}$$