In any ΔABC, prove that

(c² – a² + b²) tan A = (a² – b² + c²) tan B = (b² – c² + a²) tan C

Need to prove: (c² – a² + b²) tan A = (a² – b² + c²) tan B = (b² – c² + a²) tan C

We know,

------ (a)

Similarly, and

Therefore,

[from (a)]

Similarly,

and

Hence we can conclude comparing above equations,

(c² – a² + b²) tanA = (a² – b² + c²) tanB = (b² – c² + a²) tanC

[Proved]