Question

If $A+B+C=\pi$, show that ${\tan }^2\frac{A}{2}+{\tan }^2\frac{B}{2}+{\tan }^2\frac{C}{2}\ge 1$.

Answer 1

Given

$A+B+C=\pi$

If we take, $A=B=C=\frac{\pi }{3}$

Then,

${\tan }^2\frac{A}{2}+{\tan }^2\frac{B}{2}+{\tan }^2\frac{C}{2}$

$={\tan }^2\left(\frac{\pi }{6}\right)+{\tan }^2\left(\frac{\pi }{6}\right)+{\tan }^2\left(\frac{\pi }{6}\right)$

$=3{\tan }^2\left(\frac{\pi }{6}\right)$

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

$=1$

Hence proved that,

${\tan }^2\frac{A}{2}+{\tan }^2\frac{B}{2}+{\tan }^2\frac{C}{2}=1$.