Question

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

Answer 1

$\frac{{\tan }^2{60}^o+4{\sin }^2{45}^o+{\sin }^2{90}^o}{3{\sec }^2{30}^o+{\csc }^2{60}^o-{\cot }^2{30}^o}$
$=\frac{(\sqrt{3}{)}^2+4{\left(\frac{1}{\sqrt{2}}\right)}^2+(1{)}^2}{3\times {\left(\frac{2}{\sqrt{3}}\right)}^2+{\left(\frac{1}{\sqrt{3}}\right)}^2-(\sqrt{3}{)}^2}$
$=\frac{3+4\times \frac{1}{2}+1}{3\times \frac{4}{3}+\frac{4}{3}-3}=\frac{3+2+1}{4+\frac{4}{3}-3}$
$=\frac{6}{\frac{12+4-9}{3}}=\frac{6\times 3}{7}=\frac{18}{7}$