Question

Find value of:
$3{\cos }^2{30}^o+{\sec }^2{30}^o+2{\cos }^2{0}^o+3{\sin }^2{90}^o-{\tan }^2{60}^o$

Answer 1

We know the value of $cos{30}^o=\frac{\sqrt{3}}{2},sec{30}^o=\frac{2}{\sqrt{3}},cos{0}^o=1,sin{90}^o=1,tan{60}^o=\sqrt{3}$. Substituting these values in the expression given in the question, we obtain,

$3.\frac{3}{4}+\frac{4}{3}+2+3-3=\frac{67}{12}$