Question

If $A$, $B$, $C$ be the angles of a triangles and
$\left|\begin{array}{ccc}\cos (A-B)& \cos (B-C)& \cos (C-A)\\ \cos (A+B)& \cos (B+C)& \cos (C+A)\\ \sin (A+B)& \sin (B+C)& \sin (C+A)\end{array}\right|=0$,
then prove that the triangle is an isosceles triangle.

Answer 1

Expand the determinant w.r.t. Its row
$\Delta =\cos (A-B)\sin (C+A-B-C)+...+...$
$=\cos (A-B)\sin (A-B)+...+...$
$=\frac{1}{2}\left[\sin (2A-2B)+\sin (2B-2C)+\sin (2C-2A)\right]$
$=\frac{1}{2}(-4)\sin (A-B)\sin (B-C)\sin (C-A)$,
$=-2\sin (A-B)\sin (B-C)\sin (C-A)=0$
$\Rightarrow$ either $A=B$ or $B=C$ or $C=A$
i.e. if $A,B,C$ be the angles of a triangle then the triangle must be isosceles.