Question

$\text{In}\Delta \text{ABC, it is being given that}$
$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ \cot \frac{A}{2}& \cot \frac{B}{2}& \cot \frac{C}{2}\\ \tan \frac{B}{2}+\tan \frac{C}{2}& \tan \frac{A}{2}+\tan \frac{C}{2}& \tan \frac{A}{2}+\tan \frac{B}{2}\end{array}\right|=0,$
then the triangle must be

• A. $\text{equilateral}$
• B. $\text{isosceles}$
• C. $\text{obtuse angled}$
• D. $\text{right angled}$

Answer 1

The correct option is B $\text{isosceles}$

$\left|\begin{array}{ccc}1& 1& 1\\ \cot \frac{A}{2}& \cot \frac{B}{2}& \cot \frac{C}{2}\\ \tan \frac{B}{2}+\tan \frac{C}{2}& \tan \frac{C}{2}+\tan \frac{A}{2}& \tan \frac{A}{2}+\tan \frac{B}{2}\end{array}\right|=0$

Operating $C_1\to C_1-C_2$ and $C_2\to C_2-C_3$

$\left|\begin{array}{ccc}0& 0& 1\\ \frac{\tan \frac{B}{2}-\tan \frac{A}{2}}{\tan \frac{A}{2}\tan \frac{B}{2}}& \frac{\tan \frac{C}{2}-\tan \frac{B}{2}}{\tan \frac{B}{2}\tan \frac{C}{2}}& \cot \frac{C}{2}\\ \tan \frac{B}{2}-\tan \frac{A}{2}& \tan \frac{C}{2}-\tan \frac{B}{2}& \tan \frac{A}{2}+\tan \frac{B}{2}\end{array}\right|=0$

$\Rightarrow (\tan \frac{B}{2}-\tan \frac{A}{2})(\tan \frac{C}{2}-\tan \frac{B}{2})\times \left|\begin{array}{ccc}0& 0& 1\\ \frac{1}{\tan \frac{A}{2}\tan \frac{B}{2}}& \frac{1}{\tan \frac{B}{2}\tan \frac{C}{2}}& \cot \frac{C}{2}\\ 1& 1& \tan \frac{A}{2}+\tan \frac{B}{2}\end{array}\right|=0$

Expanding along $R_1$ we get

$(\tan \frac{B}{2}-\tan \frac{A}{2})(\tan \frac{C}{2}-\tan \frac{B}{2})(\tan \frac{C}{2}-\tan \frac{A}{2})=0$

$\Rightarrow A=B$ or $B=C$ or $C=A$

$\Rightarrow$The triangle must be an isosceles triangle