IMO 1967 LL MON35

Origin: MON

Problem

Prove the identity n  k=0 n k   tan x 2k 1 + 2k
1 −tan2(x/2) k = sec2n x 2 + secn x.

Solution

The given sum can be rewritten as n  k=0 n k   tan2 x k + n  k=0 n k   2 tan2 x 1 −tan2 x k . Since 2 tan2(x/2) 1−tan2(x/2) = 1−cos x cos x , the above sum is transformed using the bino- mial formula into  1 + tan2 x n +  1 + 1 −cos x cos x n = sec2n x 2 + secn x.