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