Prove-sec-2-x-4-n-0-1-2n-1-pi-2x-2-1-2n-1-pi-2x-2-

Question Number 142595 by qaz last updated on 02/Jun/21

Prove :: \sec^{2} x = 4 \underset{n = 0}{\sum^{\infty}} {\frac{1}{{[(2 n + 1) π - 2 x]}^{2}} + \frac{1}{{[(2 n + 1) π + 2 x]}^{2}}}

Answered by Dwaipayan Shikari last updated on 02/Jun/21

t a n x = \frac{1}{\frac{π}{2} - x} - \frac{1}{\frac{π}{2} + x} + \frac{1}{\frac{3 π}{2} - x} - \frac{1}{\frac{3 π}{2} + x} + . .

= \underset{n = 0}{\sum^{\infty}} \frac{1}{(2 n + 1) \frac{π}{2} - x} - \frac{1}{(2 n + 1) \frac{π}{2} + x}

D . b . s . w . r . t x

{s e c}^{2} x = 4 \underset{n = 0}{\sum^{\infty}} \frac{1}{{((2 n + 1) π - 2 x)}^{2}} + \frac{1}{{((2 n + 1) π + 2 x)}^{2}}