Q. 64.3( 13 Votes )
In a ∆ABC, prove that :
i. cos (A + B) + cos C = 0
ii. 
iii. 
Answer :
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
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