How would one prove that: arctan x + arctan( (1)/(x) ) = (\pi )/(2) what would happen if x < 0?

Question

How would one prove that:


`arctan x + arctan( (1)/(x) ) = (\pi )/(2)`

what would happen if x < 0?

Answer 1

Answer: If x<0, the identity does not always hold.

Explanation:

If
`0 < \theta < (\pi)/(2)`, let
`\arctan x=\theta`, implying
`x=\tan \theta`, and vice versa.

We know that
`\tan \left((\pi)/(2)-\theta \right)=\cot \theta=(1)/(\tan \theta)`, so this means that if we let
`\tan \theta=x`

`\tan \left((\pi)/(2) -\theta \right)=(1)/(x)\\(\pi)/(2)-\theta=\arctan \left((1)/(x) \right)\\\theta+\arctan \left((1)/(x) \right)=(\pi)/(2)`

Substituting back, we obtain that
`\arctan (x)+\arctan \left((1)/(x) \right)=(\pi)/(2)`, as required.