In a ∆ABC, prove that :

i. cos (A + B) + cos C = 0

ii.

iii.

We know that in ΔABC, A + B + C = π

(i) Here A + B = π – C

LHS = cos (A + B) + cos C

= cos (π – C) + cos C

We know that cos (π – C) = -cos C

= -cos C + cos C

= 0

= RHS

Hence proved.

(ii) ⇒ A + B = π – C

LHS

We know that

= RHS

Hence proved.

(iii)

⇒ A + B = π – C

LHS

We know that

= RHS

Hence proved