Let f(x) = arctan x + arccot x.
Find f(0) + f(1) + f(sqrt{2}) + f(sqrt{3}).
+150 For all values of x in the function \(f(x) = tan^{-1}(x) + cot^{-1}(x)\), \(f(x)\) will always be equal to \(\pi/2\)
Therefore, \(f(0) + f(1) + f(\sqrt{2}) + f(\sqrt{3}) = \frac{\pi}{2} + \frac{\pi}{2} + \frac{\pi}{2} + \frac{\pi}{2} = 2\pi\)
+150 That's actually a really good question.
I actually only noticed that property when graphing \(f(x)\) into Desmos out of curiosity, as seen here: https://www.desmos.com/calculator/0f7lengu0o
To answer your question, the Cofunction Identity.
The basic cofunction identity regarding tangent and cotangent state:
\(tan(\theta) = cot(\frac{\pi}{2}-\theta)\) and also \(cot(\theta) = tan(\frac{\pi}{2}-\theta)\)
Simply put into words, the tangent of an angle will equal the cotangent of the angle's complement.
It's really hard to put into words, but to me, this makes the sum of the inverses equaling \(\frac{\pi}{2}\) make sense.
Something else to note is the end behaviour, or what some consider range in this case, of both of these graphs.
For \(g(x) = tan^{-1}(x)\), we have \(\lim_{x\rightarrow \infty}g(x) = \frac{\pi}{2}\) and \(\lim_{x\rightarrow -\infty}g(x) = -\frac{\pi}{2}\)
For \(h(x) = cot^{-1}(x)\), we have \(\lim_{x\rightarrow \infty}h(x) = 0\) and \(\lim_{x\rightarrow -\infty}h(x) = \pi\).
In other words, both of these functions are defined for all values of x. However, the y values of arctan are bounded between \((-\frac{\pi}{2},\frac{\pi}{2})\) and the y values of arccot are bounded between \((0,\pi)\).
Summing these end behaviours together is where we note the behaviour of your given \(f(x)\).
If \(f(x) = tan^{-1}(x) + cot^{-1}(x) = g(x) + h(x)\), then
\(\lim_{x\rightarrow \infty}f(x) = \lim_{x\rightarrow \infty}g(x) + \lim_{x\rightarrow \infty}h(x) = \frac{\pi}{2} + 0 = \frac{\pi}{2}\)
and \(\lim_{x\rightarrow -\infty}f(x) = \lim_{x\rightarrow -\infty}g(x) + \lim_{x\rightarrow -\infty}h(x) = -\frac{\pi}{2} + \pi = \frac{\pi}{2}\)
There may be other ways to show why \(f(x)\) behaves this way, but these are the ones that immediately came to mind and working algebraically working with inverse functions is rough lol.
