Simplify by introducing $ax=y$. You now need to solve $\tan(y)=b\sec\left(\frac ya\right),\frac ya>0$. If you expand using the $e^{ix}$ definitions of $\tan,\sec$, substitute $y=i w$ and then $w=\ln(c)$ you get this equation, a power function equation which has non-elementary inverses. What have you tried?

I initially tried doing an average and even approximating, but no success. I then thought about $e^{ix}$ definition but ended up in a mess.

What about approaching it using Taylor series, Fourier series or similar?

You can use an inversion theorem or use radicals for small $a$. However, I am not used to using Fourier series.

Me neither just saying. Could you elaborate on the radicals and small $a$ part? Also, I'll look into this inversion theorem.

Maybe not radicals, but you need to find the inverse of a polynomial like here for $a=4$. Then plug in $x=0$.

Why would you plug in $x = 0$?

Is this the chat you're talking about? Sorry, I am new to this.

@JoanSGF When did I mention a chat? That probably was an automatic message since we have been sending many messages. Since you want to find the value such that $f(y)=x$, one solution is that $y=f^{-1}(x)$, to find where $f(y)=0$, plug in $x=0$ to find that $f(y)=0\implies y=f^{-1}(0)$. The MathJax does not show up in the chat.

Hello @ClaudeLeibovici

Yes I notices and thanks for the explanation I now get it.

@TymaGaidash Thats's basically the solution right? WolframAlpha gives it in form of # functions or whatever are those.

I'll find the inverse function tomorrow morning (at least in my time zone). I'll go to sleep.

Hello @TymaGaidash ! How are you ?

@ClaudeLeibovici I am doing well. How are you? @JoanSGF The “# functions” are actually “#1”s which are supposed to be a variable. For example, “#1 ^2 b^2” should be seen as $x^2 b^2$. The “Root” operator is here

I'll be looking for both: the Lagrange's inverse and the one you sent me on WolframAlpha

What would $w$ be in $z=f(w)$ for the Lagrangian?

Anyways I figured out on myown

For the Lagrange inverse, since it required $f'(k)≠0$, I used $k=0.05$ because when sliding the $a$ and $b$ parameters, it almost covered all y values between 0 and $\pi/2$ for the equation of $\frac{y}{a}>0$; and so, I used this small $k$ for which $a$ and $b$ are unusual, this way working for most $a$ and $b$.

Now I should just make a Python code which can compute this long formula

No success unfortunately :( : I couldn't write the code nor reduce the g_n component of the formula of the Lagrangian inverse. Furthermore, it looks like WolframAlpha can find inverses for that tan and sec equation for a values bigger than 1, but since a is between 0 and 1, then it cannot find any inverses. So non of these methods worked.

The hard part about the code is the n-1 th derivative and its limit

@JoanSGF There was a third user in this room who is experienced with giving approximations. Maybe you can ask @ClaudeLeibovici for help? The limit in the inversion theorem is just to evaluate the derivatives at you chosen value of $k$. Finally, you can write what you have tried in your question to make it more detailed as mentioned here. Hopefully, it will attract other users to help. Thanks

Do I ask him personally or wait for him to join? What do I mention? The inversion theorem and the failed Pythoin code?