n varying from 1 to N, show that the product

tan((n*pi)/2N+1) = (2N+1)^(1/2)

Daer Aritra,

Tanx can be expanded using Taylors expansion theorem to:

x + x^3/3 + 2x^5/15 + 17x^7/315 + ...

then we substitute x by n*pi;

The given series can be solved by taking N common and applying the limits that n varies from 1 to N

(Note- we apply limits 1-N not 0-N)

Thus the LHS reduces to ((2N+1)^3/2 )/ 2N+1,

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