Question

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

Answer 1

$\frac{{\tan }^{2}60°+4{\cos }^{2}45°+3{\sec }^{2}30°+5{\cos }^{2}90°}{cosec30°+\sec 60°-{\cot }^{2}30°}$

$\Rightarrow \frac{{\left(\sqrt{3}\right)}^2+4\times {\left(\frac{1}{\sqrt{2}}\right)}^2+3{\left(\frac{2}{\sqrt{3}}\right)}^2+0}{2+2-{\left(\sqrt{3}\right)}^2}=\frac{3+2+4}{4-3}=9$

Hence, the answer is $9.$