For any Δ ABC show that - c (a cos B – b cos A) = a2 – b2
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:
c (a cos B – b cos A) = a2 – b2
As LHS contain ca cos B and cb cos A which can be obtained from cosine formulae.
∴ From cosine formula we have:
Cos A =
⇒ bc cos A = …..eqn 1
And Cos B =
⇒ ac cos B = ……eqn 2
Subtracting eqn 1 from eqn 2:
ac cos B - bc cos A =
⇒ ac cos B - bc cos A =
∴ c (a cos B – b cos A ) = a2 – b2 …proved