Question

Find the value of:
${\cos }^2{60}^o+{\sec }^2{30}^o$

Answer 1

${\cos }^2{60}^{\circ }+{\sec }^2{30}^{\circ }$

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

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

$=\frac{3+16}{12}=\frac{19}{12}$