Question

In a $\Delta ABC$ , if $\left|\begin{array}{ccc}1& a& b\\ 1& c& a\\ 1& b& c\end{array}\right|=0$ , 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)

$\left|\begin{array}{ccc}1& a& b\\ 1& c& a\\ 1& b& c\end{array}\right|=0$
$R_2\to R_2-R_1,R_3\to R_3-R_2$
$\left|\begin{array}{ccc}1& a& b\\ 0& c-a& a-b\\ 0& b-c& c-a\end{array}\right|=0$
$\Rightarrow 1((c-a-(a-b)(b-c))=0$
$\Rightarrow c^2+a^2-2ca-ab+ac+b^2-bc=0$
$\Rightarrow a^2+b^2+c^2-ab-bc-ca=0$
$\Rightarrow \frac{1}{2}\left[{\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2\right]=0$
$\therefore a=b,b=c,c=a\Rightarrow a=b=c$
$\therefore \angle A=\angle B=\angle C=60^{\circ }$
$\therefore {\left(tan\right)}^2\left(\frac{A}{2}\right)+{\left(tan\right)}^2\left(\frac{B}{2}\right)+{\left(tan\right)}^2\left(\frac{C}{2}\right)$
$=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$