Question

${\cos }^2{30}^{\circ }{\cos }^2{45}^{\circ }+4{\sec }^2{60}^{\circ }+\frac{1}{2}{\cos }^2{90}^{\circ }-2{\tan }^2{60}^{\circ }=$

• A. $\frac{73}{8}$
• B. $\frac{75}{8}$
• C. $\frac{81}{8}$
• D. $\frac{83}{8}$

Answer 1

Answer: (D)

$\frac{83}{8}$

Taking ${\cos }^2{30}^{\circ }{\cos }^2{45}^{\circ }+4{\sec }^2{60}^{\circ }+\frac{1}{2}{\cos }^2{90}^{\circ }-2{\tan }^2{60}^{\circ }$

substitute the values of $\cos {30}^{\circ }$, $\cos {45}^{\circ }$, $\cos {90}^{\circ }$, $\sec {60}^{\circ }$, $\tan {60}^{\circ }$

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

$=(\frac{3}{4}\times \frac{1}{2})+16-6$

$=\frac{3}{8}+10$

$=\frac{3+80}{8}$

$=\frac{83}{8}$