Question

If $\tan A+\tan B=2$, then what is the value of ${\tan }^{2}A+{\tan }^{2}B$?

Answer 1

${\tan }^{2}A+{\tan }^{2}B={(\tan A+\tan B)}^2-2\tan A\tan B$ ........$\left(1\right)$

We know $A.M\ge G.M$

$\Rightarrow \frac{\tan A+\tan B}{2}\ge {\left(\tan A\tan B\right)}^{\frac{1}{2}}$

Squaring both sides,we get

$\Rightarrow \frac{{(\tan A+\tan B)}^2}{4}\ge \tan A\tan B$

$\Rightarrow {(\tan A+\tan B)}^2\ge 4\tan A\tan B$

$\Rightarrow {\left(2\right)}^2\ge 4\tan A\tan B$

$\Rightarrow 4\tan A\tan B\le 4$

$\Rightarrow \tan A\tan B\le 1$ .........$\left(2\right)$

$\Rightarrow {\tan }^{2}A+{\tan }^{2}B+2\tan A\tan B\ge 4\tan A\tan B$ from $\left(1\right)$

$\Rightarrow {\tan }^{2}A+{\tan }^{2}B\ge 2\tan A\tan B$

$\therefore ,{\tan }^{2}A+{\tan }^{2}B\ge 2\times 1=2$ from $\left(2\right)$

Hence ${\tan }^{2}A+{\tan }^{2}B\ge 2$