NIMCET 2021 — Mathematics PYQ

Proof that $k\ge 1$ 
Step 1: Use the identity for sum of products of half-angle tangents 
In any triangle $△ABC$, the sum of the products of the tangents of the half-angles is known to be equal to $1$. This identity is expressed as: $\tan \left(\frac{A}{2}\right)\tan \left(\frac{B}{2}\right)+\tan \left(\frac{B}{2}\right)\tan \left(\frac{C}{2}\right)+\tan \left(\frac{C}{2}\right)\tan \left(\frac{A}{2}\right)=1$. 
Step 2: Apply the AM-GM inequality 
For any three positive real numbers $x,y,z$, the arithmetic mean-geometric mean (AM-GM) inequality states that $\frac{x+y+z}{3}\ge \sqrt{xyz}$. A specific application of this inequality for squares of real numbers is that for any real numbers $a,b,c$, the following inequality holds: $a^2+b^2+c^2\ge ab+bc+ca$.   
Step 3: Substitute and conclude 
Let $a=\tan \left(\frac{A}{2}\right)$, $b=\tan \left(\frac{B}{2}\right)$, and $c=\tan \left(\frac{C}{2}\right)$. Since $A,B,C$ are angles of a triangle, $\frac{A}{2},\frac{B}{2},\frac{C}{2}$ are acute angles, and thus their tangents are positive. Substituting these into the inequality from Step 2, it is obtained that: ${\tan }^2\left(\frac{A}{2}\right)+{\tan }^2\left(\frac{B}{2}\right)+{\tan }^2\left(\frac{C}{2}\right)\ge \tan \left(\frac{A}{2}\right)\tan \left(\frac{B}{2}\right)+\tan \left(\frac{B}{2}\right)\tan \left(\frac{C}{2}\right)+\tan \left(\frac{C}{2}\right)\tan \left(\frac{A}{2}\right)$. From Step 1, the right-hand side of this inequality is equal to $1$. Therefore, it is concluded that: ${\tan }^2\left(\frac{A}{2}\right)+{\tan }^2\left(\frac{B}{2}\right)+{\tan }^2\left(\frac{C}{2}\right)\ge 1$. Given that $k={\tan }^2\left(\frac{A}{2}\right)+{\tan }^2\left(\frac{B}{2}\right)+{\tan }^2\left(\frac{C}{2}\right)$, it follows that $k\ge 1$.   
Final Answer 
The final answer is \boxed{\text{≥1}}.