Hello, damanwithdaplan!

If we look at it the right way, we don't need fancy formulas.

Determine: .$\displaystyle \sin\left[\arctan p+\arctan\frac{1}{p}\right]\;\text{ if }p>0 $

Let: .$\displaystyle \begin{Bmatrix} \alpha &=& \arctan p \\ \beta &=& \arctan\frac{1}{p} \end{Bmatrix}$ .[1]

The problem becomes: .$\displaystyle \sin(\alpha + \beta)$

From [1], we have: .$\displaystyle \begin{Bmatrix}\tan\alpha &=& p \\ \tan\beta &=&\frac{1}{p}\end{Bmatrix}$

And we have this right triangle: Code:

*
* β|
* |
* | p
* |
* α |
* - - - - - - - - *
1

Get it?

The two angles are from the *same right triangle!*

. . That is, $\displaystyle \alpha + \beta \:=\:90^o$

Therefore: .$\displaystyle \sin(\alpha + \beta) \;=\;\sin(90^o) \;=\;1$