Question -
Answer -
Let us consider LHS:
(c2┬атАУ a2┬а+ b2),(a2┬атАУ b2┬а+ c2), (b2┬атАУc2┬а+ a2)
We know sine rule in ╬Ф ABC
As LHS contain┬а(c2┬атАУ a2┬а+b2), (a2┬атАУ b2┬а+ c2)and┬а(b2┬атАУ c2┬а+ a2), which canbe obtained from cosine formulae.
From cosine formula we have:
Cos A = (b2┬а+ c2┬атАУ a2)/2bc
2bc cos A = (b2┬а+ c2┬атАУa2)
Let us multiply both the sides by tan A we get,
2bc cos A tan A = (b2┬а+ c2┬атАУa2) tan A
2bc cos A (sin A/cos A) = (b2┬а+ c2┬атАУa2) tan A
2bc sin A = (b2┬а+ c2┬атАУa2) tan A тАж (i)
Cos B = (a2┬а+ c2┬атАУ b2)/2ac
2ac cos B = (a2┬а+ c2┬атАУb2)
Let us multiply both the sides by tan B we get,
2ac cos B tan B = (a2┬а+ c2┬атАУb2) tan B
2ac cos B (sin B/cos B) = (a2┬а+ c2┬атАУb2) tan B
2ac sin B = (a2┬а+ c2┬атАУb2) tan B тАж (ii)
Cos C = (a2┬а+ b2┬атАУ c2)/2ab
2ab cos C = (a2┬а+ b2┬атАУc2)
Let us multiply both the sides by tan C we get,
2ab cos C tan C = (a2┬а+ b2┬атАУc2) tan C
2ab cos C (sin C/cos C) = (a2┬а+ b2┬атАУc2) tan C
2ab sin C = (a2┬а+ b2┬атАУc2) tan C тАж (iii)
As we are observing that sin terms are being involvedso letтАЩs 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
тИ┤┬аFrom┬аequation(i), (ii) and (iii) we have,
(c2┬атАУ a2┬а+ b2)tan A = (a2┬атАУ b2┬а+ c2) tan B = (b2┬атАУc2┬а+ a2) tan C
Hence proved.