Question

Evaluate: $\frac{{\tan }^2{60}^o+4{\cos }^2{45}^o+3{\sec }^2{30}^o+5{\cos }^2{90}^o}{co\sec {30}^o+\sec {60}^o-{\cot }^2{30}^o}$.

Answer 1

$\frac{ta{n}^2{60}^0+4co{s}^2{45}^0+3se{c}^2{30}^0+5co{s}^2{90}^0}{cosec{30}^0+sec{60}^0-co{t}^2{30}^0}=\frac{3+2+4+0}{2+2-3}=9$