Question

If ${\cos }^2{60}^{\circ }+{\sec }^2{30}^{\circ }+{\tan }^2{45}^{\circ }$ is $m\frac{7}{12}$, then $m$ is:

Answer 1

${\cos }^2{60}^{\circ }+{\sec }^2{30}^{\circ }+{\tan }^{2}45{}^0$
= $(\frac{1}{2}{)}^2+(\frac{2}{\sqrt{3}}{)}^2+1^2$
= $\frac{1}{4}+\frac{4}{3}+1$

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

= $\frac{31}{12}$

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