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]


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