A) equilateral
B) isosceles
C) obtuse angled
D) none of these
Correct Answer: B
Solution :
[b] Applying \[{{C}_{1}}\to {{C}_{1}}-{{C}_{2}},{{C}_{2}}\to {{C}_{2}}-{{C}_{3}}\], we get\[\Delta =\left| \begin{matrix} 0 & 0 & 1 \\ \cot \frac{A}{2}-\cot \frac{B}{2} & \cot \frac{B}{2}-\cot \frac{C}{2} & \cot \frac{C}{2} \\ \tan \frac{B}{2}-\tan \frac{A}{2} & \tan \frac{C}{2}-\tan \frac{B}{2} & \tan \frac{A}{2}+\tan \frac{B}{2} \\ \end{matrix} \right|\] \[\Delta =\left| \begin{matrix} 0 & 0 & 1 \\ \cot \frac{A}{2}-\cot \frac{B}{2} & \cot \frac{B}{2}-\cot \frac{C}{2} & \cot \frac{C}{2} \\ \frac{\cot \frac{A}{2}-\cot \frac{B}{2}}{\cot \frac{A}{2}\cot \frac{B}{2}} & \frac{\cot \frac{B}{2}-\cot \frac{C}{2}}{\cot \frac{B}{2}\cot \frac{C}{2}} & \tan \frac{A}{2}+\tan \frac{B}{2} \\ \end{matrix} \right|\] \[=\left( \cot \frac{A}{2}-\cot \frac{B}{2} \right)\left( \cot \frac{B}{2}-\cot \frac{C}{2} \right)\] \[\times \left| \begin{matrix} 0 & 0 & 1 \\ 1 & 1 & \cot \frac{C}{2} \\ \tan \frac{A}{2}\tan \frac{B}{2} & \tan \frac{B}{2}\tan \frac{C}{2} & \tan \frac{A}{2}\tan \frac{B}{2} \\ \end{matrix} \right|\]\[=\left( \cot \frac{A}{2}-\cot \frac{B}{2} \right)\left( \cot \frac{B}{2}-\cot \frac{C}{2} \right)\] \[\left( \tan \frac{C}{2}-\tan \frac{A}{2} \right)\tan \frac{B}{2}\] Since\[\Delta =0\], we have \[\cot \frac{A}{2}=\cot \frac{B}{2}\] Or \[\cot \frac{B}{2}=\cot \frac{C}{2}\] \[\tan \frac{A}{2}=\tan \frac{C}{2}\] Hence, the triangle is definitely isosceles.
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