For any triangle ABC, prove that`(b^2 - c^2) cotA + (c^2 - a^2) cotB + (a^2 - b^2) cotC = 0`

Explanation

Since `a = 2R sin A, b = 2R sin B, and c = 2R sin C`, we have
`(b^(2) -c^(2)) cot A = 4R^(2) (sin^(2) B - sin^(2) C) cot A`
`=4R^(2) sin(B + C) sin (B - C) cot A`
`= 4R^(2) sin A sin (B - C) (cos A)/(sin A)`
`= -4R^(2) sin (B - C) cos (B + C) " " ( :' cos A = - cos (B + C))`
`= -2R^(2) [2 sin (B - C) cos (B + C)]`
`= -2R^(2) [sin 2B - sin 2C]`(i)
Similarly, `(c^(2) -a^(2)) cot B = - 2R^(2) [sin 2 C - sin 2A]` (ii)
and `(a^(2) -b^(2)) cot C = -2R^(2) [sin 2A - sin 2B]` (iii)
Adding Eqs. (i), (ii), and (iii), we get
`(b^(2) - c^(2)) cot A + (c^(2) - a^(2)) cot B + (a^(2) - b^(2)) cot C = 0`