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]