Answer
$sec~72^{\circ} = \sqrt{5}+1$
Work Step by Step
$cos~x = sin(90^{\circ}-x)$
We can find the value of $cos~72^{\circ}$:
$cos~72^{\circ} = sin(90^{\circ}-72^{\circ})$
$cos~72^{\circ} = sin(18^{\circ})$
$cos~72^{\circ} = \frac{\sqrt{5}-1}{4}$
We can find the value of $sec~72^{\circ}$:
$sec~72^{\circ} = \frac{1}{cos~(72^{\circ})}$
$sec~72^{\circ} = \frac{1}{\frac{\sqrt{5}-1}{4}}$
$sec~72^{\circ} = \frac{4}{\sqrt{5}-1}$
$sec~72^{\circ} = \frac{4}{\sqrt{5}-1}~\frac{\sqrt{5}+1}{\sqrt{5}+1}$
$sec~72^{\circ} = \frac{4~(\sqrt{5}+1)}{4}$
$sec~72^{\circ} = \sqrt{5}+1$
