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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 $(-3,+\infty)$.

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    $\begingroup$ 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 $(a,b)$. $\endgroup$ – Scientifica Oct 7 '18 at 20:09
  • $\begingroup$ 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 $a < x < b)$. $\endgroup$ – bjcolby15 Oct 7 '18 at 23:48
  • $\begingroup$ Umm I thought everyone used this notation thanks for the info $\endgroup$ – AymaneLazarus Oct 10 '18 at 23:10
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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$$f^{-1}(x)=\arctan\left(\sqrt3-\sqrt{x+3}\right).$$

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  • $\begingroup$ If F=u∘v then F−1=u−1∘v-1? I'm sorry but I'm stilla highschooler so I'm fairly ignorant $\endgroup$ – AymaneLazarus Oct 7 '18 at 17:26
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    $\begingroup$ If $f=g\circ h$, then $f^{-1}=h^{-1}\circ g^{-1}$. $\endgroup$ – José Carlos Santos Oct 7 '18 at 17:28
  • $\begingroup$ Hmm this is quiet useful, but this isn't included in our program,anyways thanks for the insight $\endgroup$ – AymaneLazarus Oct 7 '18 at 17:29
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    $\begingroup$ Note that\begin{align}f(x)=y&\iff p\bigl(\tan(x)\bigr)=y\\&\iff\tan x=p^{-1}(y)\\&\iff x=\arctan\bigl(p^{-1}(y)\bigr).\end{align} $\endgroup$ – José Carlos Santos Oct 7 '18 at 17:32

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