Question

Find the value of: $\sqrt{\frac{1-{\sin }^2{60}^0}{1-{\cos }^2{60}^0}}$ is $\frac{1}{\sqrt{m}}$, m is

Answer 1

$\sqrt{\frac{1-{\sin }^2{60}^0}{1-{\cos }^2{60}^0}}$

= $\sqrt{\frac{{\cos }^2{60}^{\circ }}{{\sin }^2{60}^{\circ }}}$

= $\frac{\cos {60}^{\circ }}{\sin {60}^{\circ }}$

= $\cot {60}^{\circ }$

= $\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{m}}$

thus, $\sqrt{m}=\sqrt{3}$

$m=3$