Question -
Answer -
Let us consider LHS:
(c2 – a2 + b2),(a2 – b2 + c2), (b2 –c2 + a2)
We know sine rule in Δ ABC
As LHS contain (c2 – a2 +b2), (a2 – b2 + c2)and (b2 – c2 + a2), which canbe obtained from cosine formulae.
From cosine formula we have:
Cos A = (b2 + c2 – a2)/2bc
2bc cos A = (b2 + c2 –a2)
Let us multiply both the sides by tan A we get,
2bc cos A tan A = (b2 + c2 –a2) tan A
2bc cos A (sin A/cos A) = (b2 + c2 –a2) tan A
2bc sin A = (b2 + c2 –a2) tan A … (i)
Cos B = (a2 + c2 – b2)/2ac
2ac cos B = (a2 + c2 –b2)
Let us multiply both the sides by tan B we get,
2ac cos B tan B = (a2 + c2 –b2) tan B
2ac cos B (sin B/cos B) = (a2 + c2 –b2) tan B
2ac sin B = (a2 + c2 –b2) tan B … (ii)
Cos C = (a2 + b2 – c2)/2ab
2ab cos C = (a2 + b2 –c2)
Let us multiply both the sides by tan C we get,
2ab cos C tan C = (a2 + b2 –c2) tan C
2ab cos C (sin C/cos C) = (a2 + b2 –c2) tan C
2ab sin C = (a2 + b2 –c2) tan C … (iii)
As we are observing that sin terms are being involvedso let’s use sine formula.
From sine formula we have,
Let us multiply abc to each of the expression we get,
bc sin A = ac sin B = ab sin C
2bc sin A = 2ac sin B = 2ab sin C
∴ From equation(i), (ii) and (iii) we have,
(c2 – a2 + b2)tan A = (a2 – b2 + c2) tan B = (b2 –c2 + a2) tan C
Hence proved.