Question
If $$A, B$$ and $$C$$ are the angles of a triangle and \[\left| {\begin{array}{*{20}{c}}
1&1&1\\
{1 + \sin A}&{1 + \sin B}&{1 + \sin C}\\
{\sin A + {{\sin }^2}A}&{\sin B + {{\sin }^2}B}&{\sin C + {{\sin }^2}C}
\end{array}} \right| = 0,\] then the triangle must be
A.
Isosceles
B.
Equilateral
C.
Right-angled
D.
None of these
Answer :
Isosceles
Solution :
Using $${C_2} \to {C_2} - {C_1}$$ and $${C_3} \to {C_3} - {C_1}$$ in the given determinant, we have
\[\Delta = \left| {\begin{array}{*{20}{c}}
1&0&0\\
{1 + \sin A}&{\sin B - \sin A}&{\sin \,C - \sin \,A}\\
{\sin \,A + {{\sin }^2}A}&{{{\sin }^2}B - {{\sin }^2}A}&{{{\sin }^2}C - {{\sin }^2}A}
\end{array}} \right|\]
Now taking $$\sin \,B - \sin \,A$$ common from $$C_2$$ and $${\sin \,C - \sin \,A}$$ common from $$C_3,$$ we have
$$\Delta = \left( {\sin \,B - \sin \,A} \right)\,\left( {\sin \,C - \sin \,A} \right)$$
\[\left| {\begin{array}{*{20}{c}}
1&0&0\\
{1 + \sin A}&1&1\\
{\sin \,A + {{\sin }^2}A}&{\sin \,B + \sin \,A}&{\sin \,C + \sin \,A}
\end{array}} \right|\]
$$ = \,\left( {\sin \,B - \sin \,A} \right)\,\left( {\sin \,C - \sin \,A} \right)\,\left( {\sin \,C - \sin \,B} \right).$$
As the determinant is zero, we must have $$\sin B = \sin A{\text{ or }}\sin A{\text{ or }}\sin C = \sin A{\text{ or }}\sin C = \sin B,{\text{ that is, }}B = A{\text{ or }}C = A{\text{ or }}C = B.$$
In all three cases we will have an isosceles triangle.