Question

The value of $\frac{4}{{\tan }^2{60}^0}+\frac{1}{{\cos }^2{30}^0}-{\sin }^2{90}^0$ is equal to

Answer 1

The given expression is equal to
$=\frac{4}{{\left(\sqrt{3}\right)}^2}+\frac{1}{{\left(\frac{\sqrt{3}}{2}\right)}^2}-{\left(1\right)}^2$
$=\frac{4}{3}+\frac{4}{3}-1=\frac{5}{3}$