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\[\arctan(x)+\arctan(\frac{ 1 }{ x })=\frac{ \pi }{ 2 }\]
Consider: \[\tan (\theta)=x\]\[\tan( \frac{\pi}{2} - \theta)=\cot (\theta)=\frac{1}{x}\]
Done it thanks @Kainui
arctan x = theta tan theta = x arctan (1/x) = theta2 tan theta 2 = 1/x = 1/tan theta= cot theta
\[\theta = \tan^{-1}(x)\]\[\frac{\pi}{2}-\theta = \tan^{-1}(\frac{1}{x})\] Just arctan and add up both equations. I already had it typed out so I'll just post it anyways lol.
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