Question

In ΔABC, prove that a(b cos C – c cos B) = b² – c².

Answer 1

Taking LHS

a(b cos C – c cos B)

= ab cos C – ac cos B

= `ab((a^2 + b^2 - c^2)/(2ab)) - ac((a^2 + c^2 - b^2)/(2ac))`

= `(a^2 + b^2 - c^2 - a^2 - c^2 + b^2)/2`

= `(2b^2 - 2c^2)/2`

= b² – c² (RHS)