Question

ABC is a triangle. Then ${\tan }^2\frac{A}{2}+{\tan }^2\frac{B}{2}+{\tan }^2\frac{C}{2}$

• A. $>1$
• B. $<1$
• C. $\ge 1$
• D. $\le 1$

Answer 1

The correct option is C $\ge 1$

$A+B+C={180}^{o}C={180}^o-(A+B)\Rightarrow \frac{C}{2}=\frac{\pi }{2}-\frac{A+B}{2}\tan \left(C/2\right)=\tan (\pi /2-(A+B\left)/2\right)=\frac{1}{\tan \left(\right(A+B\left)/2\right)}=\frac{1-\tan \left(A/2\right)\tan \left(B/2\right)}{\tan \left(A/2\right)+\tan \left(B/2\right)}$

Let $a=\tan \left(A/2\right),b=\tan \left(B/2\right),c=\tan \left(C/2\right)$ all positive, the constraint becomes

$c=(1-ab)/(a+b)$

which is equivalent to $ab+bc+ca=1$

and we know that,

$a^2+b^2+c^2\ge ab+bc+ca\Rightarrow a^2+b^2+c^2\ge 1\Rightarrow {\tan }^2\left(A/2\right)+{\tan }^2\left(B/2\right)+{\tan }^2\left(C/2\right)\ge 1$

Therefore, Answer is $\ge 1$