# How do you find the value of  tan [arcsec((sqrt 5)/2)]?

$\frac{1}{2}$
we want to find $\tan \theta$ where $\theta = a r c \sec \left(\frac{\sqrt{5}}{2}\right)$ so $\sec \theta = \left(\frac{\sqrt{5}}{2}\right)$ so $\cos \theta = \frac{2}{\sqrt{5}} = \frac{a \mathrm{dj} a c e n t}{h y p o t e N e u s e}$
the opposite side is therefore $\sqrt{5 - {2}^{2}} = 1$ so $\tan \theta = \frac{1}{2}$