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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