1. Tardigrade
  2. Question
  3. Mathematics
  4. In triangle ABC, if | beginmatrix1&1&1 cot(A/2)&cot(B/2)&cot(C/2) tan(B/2)+tan(C/2)&tan(C/2)+tan(A/2)&tan(A/2)+tan(B/2) endmatrix|=0then the triangle must be

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

Determinants

Solution:

Applying $C_{1}\to C_{1}-C_{2}, C_{2}\to C_{2}-C_{3},$we get
$\Delta=\left|\begin{matrix}0&0&1\\ cot\frac{A}{2}-cot\frac{B}{2}&cot\frac{B}{2}-cot\frac{C}{2}&cot\frac{C}{2}\\ tan\frac{B}{2}-tan\frac{A}{2}&tan\frac{C}{2}-tan\frac{B}{2}&tan\frac{A}{2}+tan\frac{B}{2}\end{matrix}\right|$
$=\left|\begin{matrix}0&0&1\\ cot\frac{A}{2}-cot\frac{B}{2}&cot\frac{B}{2}-cot\frac{C}{2}&cot\frac{C}{2}\\ \frac{cot\frac{A}{2}-cot\frac{B}{2}}{cot\frac{A}{2} cot\frac{B}{2}}&\frac{cot \frac{B}{2}-cot\frac{C}{2}}{cot \frac{B}{2} cot\frac{C}{2}}&tan\frac{A}{2}+tan\frac{B}{2}\end{matrix}\right|$
$=\left(cot\frac{A}{2}-cot\frac{B}{2}\right)\left(cot\frac{B}{2}-cot\frac{C}{2}\right)$
$\times\left|\begin{matrix}0&0&1\\ 1&1&cot \frac{C}{2}\\ tan \frac{A}{2} tan\frac{B}{2}& tan \frac{B}{2} tan \frac{C}{2}& tan \frac{A}{2} tan \frac{B}{2}\end{matrix}\right|$
$=\left(cot \frac{A}{2}-cot \frac{B}{2}\right)\left(cot \frac{B}{2}-cot \frac{C}{2}\right)\left(tan \frac{C}{2}-tan \frac{A}{2}\right)tan \frac{B}{2}$
Since$\Delta=0,$we have
$cot \frac{A}{2}=cot \frac{B}{2}$ or cot $\frac{B}{2}=cot \frac{C}{2}$ or tan $\frac{A}{2}= tan \frac{C}{2}$
Hence, the triangle is definitely isosceles