How does $\pi - \arctan(\frac{8}{x}) - \arctan(\frac{13}{20-x}) = \arctan(\frac{x}{8}) + \arctan(\frac{20-x}{13})$?

Question

How does $$\pi - \arctan(\frac{8}{x}) - \arctan(\frac{13}{20-x}) = \arctan(\frac{x}{8}) + \arctan(\frac{20-x}{13})$$

Indirect way to prove this: prove it for one convenient value of x, then use $ d/dx arctan(x) = 1/(1+x^2) $ — MSalters, Aug 12 '16 at 15:29

score 4 · Accepted Answer · edited Apr 13 '17 at 12:20

Use the fact that for $t\not=0$, $\displaystyle \arctan(t)+\arctan(1/t)=\mbox{sgn}(t)\cdot\frac{\pi}{2}$.

See here for the details: $\arctan (x) + \arctan(1/x) = \frac{\pi}{2}$

Then for $x\not=0$ and $x\not=20$, $$\arctan(\frac{8}{x})+\arctan(\frac{13}{20-x})+\arctan(\frac{x}{8}) + \arctan(\frac{20-x}{13})\\ =\mbox{sgn}(x)\cdot\frac{\pi}{2}+ \mbox{sgn}(20-x)\cdot\frac{\pi}{2}=\begin{cases} \pi\quad \mbox{if $0<x<20$,}\\ 0 \quad \mbox{if $x<0$ or $x>20$.}\\ \end{cases}$$

score 2 · Answer 2 · answered Aug 12 '16 at 15:33

There's a trigonometric identity for $\arctan$: $\arctan(\alpha) \pm \arctan(\beta) = \arctan(\frac{\alpha \pm \beta}{1 \mp \alpha\beta})$.
In our case:
$$ \pi-\arctan(\frac{8}{x})-\arctan(\frac{13}{20-x})=\arctan(\frac{x}{8})+\arctan(\frac{20-x}{13})\\ \pi-\arctan(\frac{13}{20-x})-\arctan(\frac{20-x}{13})=\arctan(\frac{8}{x})+\arctan(\frac{x}{8})\\ \pi-(\arctan(\frac{13}{20-x})+\arctan(\frac{20-x}{13}))=\arctan(\frac{8}{x})+\arctan(\frac{x}{8})\\ \pi-\arctan(\frac{\frac{13}{20-x}+\frac{20-x}{13}}{1-(\frac{13}{20-x}\cdot \frac{20-x}{13})})=\arctan(\frac{\frac{8}{x}+\frac{x}{8}}{1-(\frac{8}{x}\cdot \frac{x}{8})})\\ \pi-\arctan(\frac{13^2+(20-x)^2}{0})=\arctan(\frac{8^2+x^2}{0})\\ $$ Note that the angle for $lim_{x\rightarrow \infty} \arctan(x) =\frac{\pi}{2}$ (in case $x<0$ we get $-\frac{\pi}{2})$.
hence, we get: $\pi - \frac{\pi}{2}=\frac{\pi}{2}$.

Aakash Kumar · Answer 3 · 2016-08-12T15:34:00.023

$$\frac{\pi}{2} - \arctan\left(\frac{8}{x}\right) + \frac{\pi}{2} - \arctan\left(\frac{13}{20-x}\right) = \arctan\left(\frac{x}{8}\right) + \arctan\left(\frac{20-x}{13}\right)$$ $$\mbox{arccot} \left(\frac{8}{x}\right) +\mbox{arccot}\left(\frac{13}{20-x}\right)=\arctan\left(\frac{x}{8}\right) + \arctan\left(\frac{20-x}{13}\right)$$

$$\color{red}{\boxed{ \color{black}{\mbox{arccot} \left(\frac{1}{t}\right) =\arctan t}}} $$ $\color{red}{\text{This is valid only when}\qquad t \gt 0}$

How does $\pi - \arctan(\frac{8}{x}) - \arctan(\frac{13}{20-x}) = \arctan(\frac{x}{8}) + \arctan(\frac{20-x}{13})$?

3 Answers3