A geometry problem by Aakhyat Singh

Since $\sin(b + c - a) , \sin(c + a - b) , \sin(a + b - c)$ are in arithmetic progression, then it follows that,

$\sin(a+b-c) - \sin(c + a - b) = \sin(c + a - b) - \sin(b + c - a)$

Now, for any two angles X and Y, we have the following identity,

$\sin X - \sin Y = \sin(X' + Y') - \sin(X' - Y') = 2 \cos X' \sin Y'$ , where $X' = \dfrac{1}{2} (X + Y)$ and $Y' = \dfrac{1}{2} (X - Y)$

Using this identity in the above equality,

$\sin(a+b-c) - \sin(c + a - b) = 2 \cos (a) \sin( b - c )$

And,

$\sin(c + a - b) - \sin(b + c - a)= 2 \cos(c ) \sin(a - b)$

Hence,

$\cos a \sin(b-c) = \cos c \sin(a - b)$

Expanding,

$\cos a . ( \sin b \cos c - \cos b \sin c ) = \cos c ( \sin a \cos b - \cos a \sin b )$

Now divide through by $\cos a \cos b \cos c$ , you get,

$\tan b - \tan c = \tan a - \tan b$

Negate it, to obtain,

$\tan b - \tan a = \tan c - \tan b$

Thus, $\tan a , \tan b, \tan c$ are in arithmetic progression, therefore, their reciprocals , $\cot a , \cot b , \cot c$ are in harmonic progression.