Question

If A, B and C are the angles of a triangle and
$\left|\begin{array}{ccc}1& 1& 1\\ 1+\sin A& 1+\sin B& 1+\sin C\\ \sin A+{\sin }^{2}A& \sin B+{\sin }^{2}B& \sin C+{\sin }^{2}C\end{array}\right|=0$
then the triangle must be

• A. isosceles
• B. equilateral
• C. right angled
• D. none of these

Answer 1

The correct option is A isosceles

$\left|\begin{array}{ccc}1& 1& 1\\ 1+\sin A& 1+\sin B& 1+\sin C\\ \sin A+{\sin }^{2}A& \sin B+{\sin }^{2}B& \sin C+{\sin }^{2}C\end{array}\right|=0$

Applying $C_2\to C_2-C_1,C_3\to C_3-C_1$

$\left|\begin{array}{ccc}1& 0& 0\\ 1+\sin A& \sin B-\sin A& \sin C-\sin A\\ \sin A+{\sin }^{2}A& {\sin }^{2}B-{\sin }^{2}A+\sin B-\sin A& {\sin }^{2}C-{\sin }^{2}A+\sin C-\sin A\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}1& 0& 0\\ 1+\sin A& \sin B-\sin A& \sin C-\sin A\\ \sin A+{\sin }^{2}A& {\sin }^{2}B-{\sin }^{2}A& {\sin }^{2}C-{\sin }^{2}A\end{array}\right|=0$

$\Rightarrow (\sin B-\sin A)(\sin C-\sin A)\left|\begin{array}{ccc}1& 0& 0\\ 1+\sin A& 1& 1\\ \sin A+{\sin }^{2}A& \sin B+\sin A& \sin C+\sin A\end{array}\right|=0$

$\Rightarrow (\sin B-\sin A)(\sin C-\sin A)(\sin C-\sin B)=0$

Hence

$\sin B=\sin A$ or $\sin C=\sin A$ or $\sin C=\sin B$

And from all three cases gives the isosceles triangle