Question

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

Answer 1

Evaluating the value of the given expression:

$$\begin{gathered} {\tan }^2{30}^0+{\tan }^2{60}^0+{\tan }^2{45}^0\left[\tan {45}^0=1,\tan {60}^0=\sqrt{3},\tan {30}^0=\frac{1}{\sqrt{3}}\right] \\ ={\left(\frac{1}{\sqrt{3}}\right)}^2+{\left(\sqrt{3}\right)}^2+{\left(1\right)}^2 \\ =\frac{1}{3}+3+1 \\ =\frac{1}{3}+4 \\ =\frac{1+12}{3} \\ =\frac{13}{3} \end{gathered}$$

Hence, ${\tan }^2{30}^0+{\tan }^2{60}^0+{\tan }^2{45}^0=\frac{13}{3}$.