Question -
Answer -
Let us consider LHS:
b (c cos A – a cos C)
As LHS contain bc cos A and ab cos C which can beobtained from cosine formulae.
From cosine formula we have:
Cos A = (b2 + c2 – a2)/2bc
bc cos A = (b2 + c2 –a2)/2 … (i)
Cos C = (a2 + b2 – c2)/2ab
ab cos C = (a2 + b2 –c2)/2 … (ii)
Now let us subtract equation (i) and (ii) we get,
bc cos A – ab cos C = (b2 + c2 –a2)/2 – (a2 + b2 – c2)/2
= c2 – a2
∴ b (c cos A – a cos C) =c2 – a2
Hence proved.