Question

${\tan }^2{30}^0+{\tan }^2{45}^0+{\tan }^2{60}^0=m\frac{1}{3}$
Then the value of $m$ is

Answer 1

${\tan }^2{30}^0+{\tan }^2{45}^0+{\tan }^2{60}^0$

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

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

$=\frac{1+3+9}{3}$

$=\frac{13}{3}$

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