Does the inverse trigonometric identity $\tan^{-1}{x} = \cot^{-1}{1 \over x}$ hold for all $x>0$?

Question

I differentiate both sides, and after a few steps I get that derivations are equal to each other, but I do not know what to conclude from those characteristics. Should I use Cauchy's or/and Lagrange's Theorem to prove that those functions are (not) equal for $x > 0$, or there is an easier way to check this?

Answer 1

In general, if $f'(x) = g'(x)$ for all $x\in D$ and $f(y) = g(y)$ for some $y\in D$ where $D$ is a connected subset of $\mathbb R$, then $f = g$ identically on $D$.

Answer 2

If two differentiable functions $f, g$ on an interval $I$ satisfy $f' = g'$ and $f(a) = g(a)$ for some $a \in I$, then $f = g$. So, the claim follows from your work so far and, e.g., the particular value $\arctan 1 = \frac{\pi}{4} = \operatorname{arccot} 1$.

Alternatively, one can just draw a right triangle with leg lengths $x, 1$ and write the acute angle adjacent to the side of length $1$ in two different ways using the definitions of $\arctan$ and $\operatorname{arccot}$.

Answer 3

If we switch the trig function we have to correspondingly change the co-function, holding the Pythagorean right angle triangle relation the same.

$$ \tan ^{-1}(x)= \cot^{-1} \left(\frac1x\right) = \cos ^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) = \sec^{-1}\left(\sqrt{1+x^2}\right)= \sin ^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) = \csc^{-1}\left(\frac{\sqrt{1+x^2}}{x}\right)...$$

They are identities and so hold for all values of $x$.

Unless there is an error, derivatives also should match. For example,

$$ \frac{d(tan^{-1} x)}{dx} = \frac{1}{1+x^2} $$

$$ \frac{d\left(cot^{-1} \left(\frac1x\right)\right)}{dx} = \dfrac{-1}{\left(1+\frac{1} {x^2}\right)} \cdot \dfrac{-1}{x^2} = \frac{1}{1+x^2}. $$