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

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

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

$$tanx=\frac{1}{\frac{\pi }{2}-x}-\frac{1}{\frac{\pi }{2}+x}+\frac{1}{\frac{3\pi }{2}-x}-\frac{1}{\frac{3\pi }{2}+x}+..$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(2n+1)\frac{\pi }{2}-x}-\frac{1}{(2n+1)\frac{\pi }{2}+x}$$$$D.b.s.w.r.tx$$$${sec}^{2}x=4\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{((2n+1)\pi -2x)}^2}+\frac{1}{{((2n+1)\pi +2x)}^2}$$

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