Question

Evaluate the following :
$ta{n}^2{30}^{\circ }+ta{n}^2{60}^{\circ }+ta{n}^2{45}^{\circ }$

Answer 1

Given $ta{n}^2{30}^{\circ }+ta{n}^2{60}^{\circ }+ta{n}^2{45}^{\circ }$

$={\left(\frac{1}{\sqrt{3}}\right)}^2+{\left(\sqrt{3}\right)}^2+{\left(1\right)}^2$ $[\because \tan {45}^{\circ }=1,\tan {30}^{\circ }=\frac{1}{\sqrt{3}},\tan {60}^{\circ }=\sqrt{3}]$

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

$=\left(\frac{1}{3}\right)+4$

$=\frac{1+12}{3}$

$=\frac{13}{3}$