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