# Find the inverse function of $\tan^2(x) - 2\sqrt {3} \tan(x)$

The question is pretty clear, we must find the inverse function of $${f:x \rightarrow \tan^2(x) - 2 \sqrt{3}(\tan(x))}$$

This function is a bijection from $$(-\frac {\pi}{2}+kπ, \frac {\pi}{3}+kπ)$$ to $$(-3,+\infty)$$ since it is strictly decreasing and continuous.

I've tried factorizing with tangent and playing with the equation so I can add Arctangent to eliminate tangent like with usual functions but I do not seem to be capable succeeding using this method.

P.S: This function is also a bijection from $$(-\frac {\pi}{3}+kπ, \frac {\pi}{2}+kπ)$$ to $$(-3,+\infty)$$.

• In this forum, the common notation for the open interval $\{x\in\mathbb R\mid a<x<b\}$ is $(a,b)$, not $]a,b[$. Many people wouldn't understand this french notation. Hence I modified it to $(a,b)$. – Scientifica Oct 7 '18 at 20:09
• Thanks, Scientifica! I originally edited the first part but didn't realize I was using brackets (which would imply $a \leq x \leq b$) when open intervals were appropriate (which would correctly imply $a < x < b)$. – bjcolby15 Oct 7 '18 at 23:48
• Umm I thought everyone used this notation thanks for the info – AymaneLazarus Oct 10 '18 at 23:10

Consider$$\begin{array}{rccc}p\colon&\left[-1,\sqrt3\right]&\longrightarrow&\left[1+2\sqrt3,-3\right]\\&x&\mapsto&x^2-2\sqrt3x.\end{array}$$Then $$f=p\circ\tan$$ and therefore $$f^{-1}=\tan^{-1}\circ p^{-1}=\arctan\circ p^{-1}$$. Finally, $$p^{-1}(x)=\sqrt3-\sqrt{x+3}$$ and therefore$$f^{-1}(x)=\arctan\left(\sqrt3-\sqrt{x+3}\right).$$
• If $f=g\circ h$, then $f^{-1}=h^{-1}\circ g^{-1}$. – José Carlos Santos Oct 7 '18 at 17:28