For any ΔABC, show that - b (c cos A – a cos C) = c2 – a2
Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .
Key point to solve the problem:
Idea of cosine formula in ΔABC
• Cos A =
• Cos B =
• Cos C =
As we have to prove:
b (c cos A – a cos C) = c2 – a2
As LHS contain bc cos A and ab cos C which can be obtained from cosine formulae.
∴ From cosine formula we have:
Cos A =
⇒ bc cos A = …..eqn 1
And Cos C =
⇒ ab cos C = ……eqn 2
Subtracting eqn 2 from eqn 1:
bc cos A - ab cos C = -
⇒ bc cos A - ab cos C = c2 - a2
∴ b(c cos A - a cos C) = c2 - a2 …proved