Question

In a $\Delta ABC$ , if $\begin{vmatrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{vmatrix}=0$ , then value of $\left(tan\right)^{2}\left(\frac{A}{2}\right)+\left(tan\right)^{2}\left(\frac{B}{2}\right)+\left(tan\right)^{2}\left(\frac{C}{2}\right)$ is equal to
[ Note: All symbols used have usual meaning in $\Delta ABC$ .]

• A. $-1$
• B. $0$
• C. $1$
• D. $\frac{1}{3}$

Answer 1

Answer: (C)

$\begin{vmatrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{vmatrix}=0$
$R_{2} \rightarrow R_{2}-R_{1},R_{3} \rightarrow R_{3}-R_{2}$
$\begin{vmatrix} 1 & a & b \\ 0 & c-a & a-b \\ 0 & b-c & c-a \end{vmatrix}=0$
$\Rightarrow 1((c-a-(a-b)(b-c))=0$
$\Rightarrow c^{2}+a^{2}-2ca-ab+ac+b^{2}-bc=0$
$\Rightarrow a^{2}+b^{2}+c^{2}-ab-bc-ca=0$
$\Rightarrow \frac{1}{2}\left[\left(a - b\right)^{2} + \left(b - c\right)^{2} + \left(c - a\right)^{2}\right]=0$
$\therefore a=b,b=c,c=a\Rightarrow a=b=c$
$\therefore \angle A=\angle B=\angle C=60^\circ $
$\therefore \left(tan\right)^{2}\left(\frac{A}{2}\right)+\left(tan\right)^{2}\left(\frac{B}{2}\right)+\left(tan\right)^{2}\left(\frac{C}{2}\right)$
$=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$