Answer
$\sec$ 45$^{\circ}$ = $\sqrt2$
Work Step by Step
$\sec$ 45$^{\circ}$
We must find side $x$ of the 45$^{\circ}$ - 45$^{\circ}$ Right Triangle.
Pythagorean Theorem: $c$$^{2}$ = $a$$^{2}$ + $b$$^{2}$
x$^{2}$ = 1$^{2}$ + 1$^{2}$
x$^{2}$ = 2
x = $\sqrt2$
Now consider the triangle from the perspective of a 45$^{\circ}$ angle.
Hypotenuse = $\sqrt2$
Opposite = 1
Adjacent = 1
$\csc$ 45$^{\circ}$ = $\frac{Hypotenuse}{Adjacent}$ = $\frac{\sqrt2}{1}$ = $\sqrt2$
Therefore:
$\sec$ 45$^{\circ}$ = $\sqrt2$
