Question

${\cos }^2{60}^0+{\sec }^2{30}^0+{\tan }^2{45}^0=m\frac{7}{12}$

Then $m$ is

Answer 1

${\cos }^2{60}^0+{\sec }^2{30}^0+{\tan }^2{45}^0$

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

$=\frac{1}{4}+\frac{4}{3}+1$

$=\frac{3+16}{12}+1$

$=\frac{19}{12}+1$

$=\frac{31}{12}$

$=2\frac{7}{12}$