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


5