Simplify arctan of x plus arctan of 1 over x.

Step-by-step solution

1. 1

   Fix the domain first. The term $\frac1x$ requires $x\neq 0$, so the expression is defined on $(-\infty,0)\cup(0,\infty)$ — two separate pieces, which is a hint that the answer may differ on each.

2. 2

   Case $x>0$: use complementary angles. Put $\theta=\arctan x$, so $\theta\in\left(0,\frac{\pi}{2}\right)$ and $\tan\theta=x$. Then

   $\tan\left(\frac{\pi}{2}-\theta\right)=\cot\theta=\frac1x$

   and $\frac{\pi}{2}-\theta$ also lies in $\left(0,\frac{\pi}{2}\right)$, which is inside the principal range of $\arctan$. Therefore $\arctan\frac1x=\frac{\pi}{2}-\theta$ and

   $\arctan x+\arctan\frac1x=\frac{\pi}{2}$

3. 3

   Case $x<0$: use oddness rather than repeating the work. Both $x<0$ and $\frac1x<0$, so both arctangents are negative and their sum cannot be $+\frac{\pi}{2}$. Apply the previous result to $-x>0$ and use $\arctan(-t)=-\arctan t$:

   $\arctan(-x)+\arctan\left(-\tfrac1x\right)=\frac{\pi}{2}\ \Rightarrow\ -\left(\arctan x+\arctan\tfrac1x\right)=\frac{\pi}{2}$

   so the sum equals $-\frac{\pi}{2}$.

4. 4

   See why the tangent addition formula alone is not enough. With $A=\arctan x$ and $B=\arctan\frac1x$,

   $\tan(A+B)=\frac{x+\frac1x}{1-x\cdot\frac1x}=\frac{x+\frac1x}{0}$

   which is undefined. That only tells us $A+B$ is an odd multiple of $\frac{\pi}{2}$; deciding which one requires the range argument above.

5. 5

   Combine into one formula.

   $\arctan x+\arctan\frac1x=\frac{\pi}{2}\operatorname{sgn}(x),\qquad x\neq 0$

   The function is constant on each half-line and jumps by $\pi$ across $x=0$ — it is emphatically not identically $\frac{\pi}{2}$.

6. 6

   Check numerically. At $x=2$: $\arctan 2=1.1071487$ and $\arctan 0.5=0.4636476$, summing to $1.5707963=\frac{\pi}{2}$ ✓. At $x=-2$ the two values negate and the sum is $-1.5707963=-\frac{\pi}{2}$ ✓.