Sue Liu, S5, Madras College sent in a good solution which shows that if $A,B$ and $C$ are angles in a triangle and

${\displaystyle \tan (A-B)+\tan (B-C)+\tan (C-A)=0}$

then the triangle is isosceles. Sue's proof uses the fact that the angles of the triangle $\mathrm{ABC}$ add up to 180 degrees. However it is just as easy to prove this result without using the fact about the sum of the angles of the triangle so it must be true for triangles 'living in other geometries' where the angles of triangles do not add up to 180 degrees such as Spherical Geometry.

The lines in Spherical Geometry are great circles on the surface of the sphere. By drawing lines like the lines of longitude and the equator on the earth you will soon be able to convince yourself that spherical triangles have angle sums greater than 180 degrees.

This expression gives isosceles triangles for all 3 geometries, for Euclidean Geometry where the angles of triangles add up to 180 degrees, for Spherical (also called Elliptical) Geometry where the angles of triangles add up to more than 180 degrees and for Hyperbolic Geometry where the angles of triangles add up to less than 180 degrees. So much for the 'territory' this result belongs to, now for the proof.

We start with the expression

${\displaystyle \tan (A-B)+\tan (B-C)+\tan (C-A)=0.}$

Write $X=A-C$ and $Y=B-C$, then the given expression becomes

${\displaystyle \tan (X-Y)+\tan Y+\tan -X=0.}$

This gives

${\displaystyle \tan (X-Y)=\tan X-\tan Y}$

and we know the identity

${\displaystyle \tan (X-Y)=\frac{\tan X-\tan Y}{1-\tan X\tan Y}.}$

Hence either

${\displaystyle \tan X=\tan Y\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(1)}$

or

${\displaystyle \tan X\tan Y=0\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(2)}$

In case (1) we show that the angles $X$ and $Y$ are equal.

${\displaystyle \Vert X-Y\Vert =\Vert A-B\Vert <A+B<{180}^{\circ }}$

and the tan function is periodic with period 180 degrees so $X=Y.$ This gives $A-C=B-C$ hence $A=B$, so the triangle is isosceles.

In case (2), either $\tan X=0$ or $\tan Y=0$, hence $A=C$ or $B=C$ and in all the cases the triangle is isosceles.