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

Answer: (B)

Applying $C_1\to C_1-C_2,C_2\to C_2-C_3,$we get
$\Delta =\left|\begin{array}{ccc}0& 0& 1\\ cot\frac{A}{2}-cot\frac{B}{2}& cot\frac{B}{2}-cot\frac{C}{2}& cot\frac{C}{2}\\ tan\frac{B}{2}-tan\frac{A}{2}& tan\frac{C}{2}-tan\frac{B}{2}& tan\frac{A}{2}+tan\frac{B}{2}\end{array}\right|$
$=\left|\begin{array}{ccc}0& 0& 1\\ cot\frac{A}{2}-cot\frac{B}{2}& cot\frac{B}{2}-cot\frac{C}{2}& cot\frac{C}{2}\\ \frac{cot\frac{A}{2}-cot\frac{B}{2}}{cot\frac{A}{2}cot\frac{B}{2}}& \frac{cot\frac{B}{2}-cot\frac{C}{2}}{cot\frac{B}{2}cot\frac{C}{2}}& tan\frac{A}{2}+tan\frac{B}{2}\end{array}\right|$
$=\left(cot\frac{A}{2}-cot\frac{B}{2}\right)\left(cot\frac{B}{2}-cot\frac{C}{2}\right)$
$\times \left|\begin{array}{ccc}0& 0& 1\\ 1& 1& cot\frac{C}{2}\\ tan\frac{A}{2}tan\frac{B}{2}& tan\frac{B}{2}tan\frac{C}{2}& tan\frac{A}{2}tan\frac{B}{2}\end{array}\right|$
$=\left(cot\frac{A}{2}-cot\frac{B}{2}\right)\left(cot\frac{B}{2}-cot\frac{C}{2}\right)\left(tan\frac{C}{2}-tan\frac{A}{2}\right)tan\frac{B}{2}$
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