Question

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

• A. Equilateral
• B. Isoscles
• C. Right angled
• D. Scalene

Answer 1

The correct option is A Equilateral

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

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

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

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

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

$=\cot \frac{A}{2}=\cot \frac{B}{2}or\cot \frac{B}{2}=\cot \frac{C}{2}or\tan \frac{A}{2}=\tan \frac{C}{2}$

$\therefore \Delta$ is isoceles $\Delta$