Express arctan(1/x) in terms of arctan(x), for x positive and for x negative.

Step-by-step solution

1. 1

   Note the domain. The expression $\tfrac1x$ requires $x\neq 0$, so the identity will have two branches with nothing to say at $x=0$.

2. 2

   Derive the positive branch from complementary angles. For $x>0$ let $\theta=\arctan x$, so $0<\theta<\tfrac{\pi}{2}$. In a right triangle with legs $1$ and $x$ the two acute angles are complementary, and swapping which leg is 'opposite' replaces $x$ by $\tfrac1x$:

   $\arctan\left(\frac{1}{x}\right)=\frac{\pi}{2}-\arctan x,\qquad x>0$

3. 3

   See why the negative branch needs a different constant. For $x<0$ both $\arctan x$ and $\arctan\tfrac1x$ lie in $\left(-\tfrac{\pi}{2},0\right)$, so their sum is negative and cannot be $+\tfrac{\pi}{2}$. The correct constant is the negative one:

   $\arctan\left(\frac{1}{x}\right)=-\frac{\pi}{2}-\arctan x,\qquad x<0$

4. 4

   Understand the jump. The function $\arctan x+\arctan\tfrac1x$ is constant on each side of the origin but jumps by $\pi$ across it, taking the value $\tfrac{\pi}{2}$ for $x>0$ and $-\tfrac{\pi}{2}$ for $x<0$. The jump exists because the principal range of arctan is only $\pi$ wide.

5. 5

   Check one value on each branch. At $x=2$: $\arctan\tfrac12\approx 0.46365$ and $\tfrac{\pi}{2}-\arctan 2\approx 1.57080-1.10715=0.46365$ $\checkmark$. At $x=-2$: $\arctan\left(-\tfrac12\right)\approx-0.46365$ and $-\tfrac{\pi}{2}-\arctan(-2)\approx-1.57080+1.10715=-0.46365$ $\checkmark$.