# Homework Help: Hyperbolic Trig Inverses

1. Dec 3, 2006

### ChaoticLlama

I need to determine an algebraic form for arcsech(x) = ?

So far what I've come up with is as follows:

$\L\ \begin{array}{l} y = \frac{2}{{e^x + e^{ - x} }} \\ y = \frac{2}{{e^x + e^{ - x} }}\left( {\frac{{e^x }}{{e^x }}} \right) \\ y = \frac{{2e^x }}{{e^{2x} + 1}} \\ ye^{2x} - 2e^x + y = 0 \\ \end{array}$

Need to continue solving for x, thanks for any suggestions.

Last edited: Dec 3, 2006
2. Dec 3, 2006

### HallsofIvy

Let z= e2x and you equation becomes
$yz^2- 2z+ y= 0$
Use the quadratic formula to solve for z and then 2x= ln z.

3. Dec 3, 2006

### arildno

Set $u=e^{x}$
Thus, you get the equation in u:

$$yu^{2}-2u+y=0$$

1.Can you solve that for u, remembering that u>0?
2. Solve afterwards for x!