Solve equation for t

Question

$$s = 2 \ln|\tan(t) + \sec(t)|$$

I tried to solve it and got a quadratic equation which turned out to equal $arcsin(\dfrac{-2 \pm e^s}{2(1+e^s)})$

This doesn't seem right. Any thoughts?

Answer 1

A better way to approach the solution:

$$ \begin{align} & s = 2 \ln|\tan(t) + \sec(t)| \\ &\frac{s}{2} = \ln\left|\frac{\sin(t)}{\cos(t)} + \frac{1}{\cos(t)}\right| \\ &e^\frac{s}{2} = \left|\frac{1+\sin(t)}{\cos(t)}\right| \\ &e^\frac{s}{2} = \left|\frac{[\cos(\frac{t}{2})+\sin(\frac{t}{2})]^2}{[\cos^2(\frac{t}{2})-\sin^2(\frac{t}{2})]}\right| \\ &e^\frac{s}{2} = \left|\frac{[\cos(\frac{t}{2})+\sin(\frac{t}{2})]}{[\cos(\frac{t}{2})-\sin(\frac{t}{2})]}\right| \\ &e^\frac{s}{2} = \left|\frac{[1+\tan(\frac{t}{2})]}{[1-\tan(\frac{t}{2})]}\right| \\ &e^\frac{s}{2} = \tan\left(\frac{\pi}{4}+\frac{t}{2}\right) \\ &t=2\arctan\left(e^\frac{s}{2}\right)-\frac{\pi}{2}\end{align}$$

Hope this helps.