A,B,C are interior angles of ∆ABC . Prove that cosec(A+B/2) = sec C/

1.	A,B,C are interior angles of ∆ABC . Prove that cosec(A+B/2) = sec C/2.
Answer» cosec(A+B)/2= cosec(180-C)/2. [since A+B+C=180]= cosec (180/2 - C/2)= cosec (90 - C/2)= sec C/2

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