Question

Solve $\frac{{\tan }^2{60}^o+4{\cos }^2{45}^o+3{\sec }^2{30}^o+5{\cos }^2{90}^o}{\csc {30}^o+\sec {60}^o-{\cot }^2{30}^o}$

Answer 1

$\frac{{\tan }^{2}60+4{\cos }^{2}45+3{\sec }^{2}30+5{\cos }^{2}90}{\csc 30+\sec 60-{\cot }^{2}30}$

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

$=\frac{3+2+4+0}{2+2-3}$

$=9$