Question

In triangle $ABC$, if $\left|\begin{array}{ccc}1& 1& 1\\ \cot \frac{A}{2}& \cot \frac{B}{2}& \cot \frac{C}{2}\\ \tan \frac{B}{2}+\tan \frac{C}{2}& \tan \frac{C}{2}+\tan \frac{A}{2}& \tan \frac{A}{2}+\tan \frac{B}{2}\end{array}\right|=0$, then the triangle must be

• A. equilateral
• B. isosceles
• C. obtuse angled
• D. none of these

Answer 1

The correct option is A equilateral

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