Note: In any ΔABC we define ‘a’ as the length of the side opposite to ∠A, ‘b’ as the length of the side opposite to ∠B and ‘c’ as the length of the side opposite to ∠C.

Key point to solve the problem:

Idea of projection Formula:

• c = a cos B + b cos A

• b = c cos A + a cos C

• a = c cos B + b cos C

As we have to prove:

a(cos B + cos C – 1) + b(cos C + cos A –1) + c(cos A + cos B – 1) = 0

We can observe that we all the terms present in equation to be proved are also present in expressions of projection formula ,so we have to apply the formula with slight modification -

∴ from projection formula we have-

c = a cos B + b cos A

⇒ b cos A + a cos B – c = 0 …..eqn 1

Also,

b = c cos A + a cos C

⇒ c cos A + a cos C – b = 0 ……eqn 2

Also,

a = c cos B + b cos C

⇒ c cos B + b cos C – a = 0 …..eqn 3

Adding eqn 1 ,2 and 3 –

We have,

b cos A + a cos B – c + c cos A + a cos C – b + c cos B + b cos C – a = 0

b cos A – b + b cos C + a cos B + a cos C – a + c cos A + c cos B – c = 0

⇒ b(cos A + cos C – 1) + a(cos B + cos C – 1) + c(cos A + cos B -1) = 0

Hence,

a(cos B + cos C – 1) + b(cos C + cos A –1) + c(cos A + cos B – 1) = 0

….proved