Question

det $\left[\begin{array}{ccc}1& 1& 1\\ sinA& sinB& sinC\\ si{n}^{2}A& si{n}^{2}B& si{n}^{2}C\end{array}\right]$ =

• A. $\frac{1}{8R^3}(a-b)(b-c)(c-a)$
• B. $8R^3(a-b)(b-c)(c-a)$
• C. $(a-b)(b-c)(c-a)$
• D. $\frac{1}{8R}(a-b)(b-c)(c-a)$

Answer 1

The correct option is A $\frac{1}{8R^3}(a-b)(b-c)(c-a)$

We know that
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$
So, $=\left|\begin{array}{ccc}1& 1& 1\\ \frac{a}{2R}& \frac{b}{2R}& \frac{c}{2R}\\ {\left(\frac{a}{2R}\right)}^2& {\left(\frac{b}{2R}\right)}^2& {\left(\frac{c}{2R}\right)}^2\end{array}\right|$
$=\frac{1}{8R^3}\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|$
$C_1\to C_1-C_2,C_2\to C_2-C_3$
$=\frac{1}{8R^3}\left|\begin{array}{ccc}0& 0& 1\\ a-b& b-c& c\\ a^2-b^2& b^2-c^2& c^2\end{array}\right|$
$=\frac{1}{8R^3}(a-b)(b-c)\left|\begin{array}{ccc}0& 0& 1\\ 1& 1& c\\ a+b& b+c& c^2\end{array}\right|$
$=\frac{1}{8R^3}(a-b)(b-c)(c-a)$