Question

In a triangle $ABC,$ the least value of ${\tan }^2\frac{A}{2}+{\tan }^2\frac{B}{2}+{\tan }^2\frac{C}{2}$ is

• A. $1$
• B. $\frac{1}{8}$
• C. $\sqrt{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is A $1$

Let $\tan \frac{A}{2}=x,\tan \frac{b}{2}=y,\tan \frac{c}{2}=z$

We know that $x^2+y^2+z^2-xy-yz-zx$

$=\frac{1}{2}[{(x-y)}^2+{(y-z)}^2+{(z-x)}^2]\ge 0$

$\Rightarrow x^2+y^2+z^2\ge xy+yz+zx$ ...(1)

Now, $A+B+C=\pi$

$\therefore \frac{A}{2}+\frac{B}{2}=\frac{\pi }{2}-\frac{C}{2}\Rightarrow \tan (\frac{A}{2}+\frac{B}{2})=\tan (\frac{\pi }{2}-\frac{C}{2})=\cot \frac{C}{2}$

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

$\Rightarrow xy+yz+zx=1$ ...(2)

from (1) and (2), $x^2+y^2+z^2\ge 1$

i.e., ${\tan }^2\frac{A}{2}+{\tan }^2\frac{B}{2}+{\tan }^2\frac{C}{2}\ge 1$