If $$ f'(x)>0 \,\forall x\in \mathbb R^+ \text{ where }f:\mathbb R^+\to \mathbb R $$ and $$ f(x)+\cfrac{1}{x} = f^{-1}\left(\cfrac{1}{f(x)}\right) $$ Also given that $$ f^{-1}\left(\cfrac{1}{f(x)}\right)>0$$ Then find out the value of $$\cfrac{π}{\sin^{-1}f(2)}$$ I tried some hit and trial to get f(2) but was unable to do so further I tried replacing $$f(x)\to y$$ then further substitutions but keep messing up every time.
Further please suggest better tags if needed.
After playing around for a while, I stumbled upon the trick:
Rewrite the equation as $$f\left(f(x) + \frac 1x\right) = \frac 1{f(x)}$$
First note that $f(x) < 0$ for all $x$:
let $M = f(\infty) := \lim_{x \to \infty} f(x)$. Since $f$ is strictly increasing, $f(x) < M$ for all $x$. Letting $x \to \infty$ in the formula, $f(M) = \frac 1M$ and letting $x = M$, $f\left(\frac 1M + \frac 1M\right) = M$ which cannot be if $M$ is finite and not $0$. But assuming $M = \infty$ leads to $f(\infty) = 0$ and $M > f(1) > -\infty$, so $M$ can only be $0$.
Now, let $y = f(2)$, then $$f\left(y +\frac 12\right) = f\left(\frac {2y + 1}2\right) = \frac 1y$$ Letting $x = \frac{2y+1}2$, $$f\left(\frac 1y + \frac 2{2y+1}\right) = f\left(\frac {4y+1}{y(2y+1)}\right) = \frac 1{\frac 1y} = y$$
But $f$ is injective. There can be only one value $x$ with $f(x) = y$. Therefore we must have $$\frac {4y+1}{y(2y+1)} = 2$$ $y$ will be the negative root.
You can generalize this to find a closed form for $f$.