# How do you know how to solve sec 45?

Jun 21, 2018

$\sec \left({45}^{\circ}\right) = \frac{\sqrt{2}}{1} = \sqrt{2}$

#### Explanation:

$\sec \left({45}^{\circ}\right)$

$= \frac{1}{\cos} \left({45}^{\circ}\right)$

$\sin \left({45}^{\circ}\right) = \cos \left({45}^{\circ}\right)$

we have an isosceles right triangle
where the ratio of its sides are $1 : 1 : \sqrt{2}$

thus:

$\sec \left({45}^{\circ}\right) = \frac{\sqrt{2}}{1} = \sqrt{2}$

Jun 22, 2018

$\sec 45 = \sqrt{2}$

#### Explanation:

We know that $\sec x$ is defined as $\frac{1}{\cos} x$, so we are essentially being asked to find

$\frac{1}{\cos} 45$

On the unit circle, the coordinates for ${45}^{\circ}$ are $\left(\frac{\sqrt{2}}{2} , \frac{\sqrt{2}}{2}\right)$, where the $x$-coordinate is the value for cosine.

This means

$\frac{1}{\cos} 45 = \frac{1}{\frac{\sqrt{2}}{2}}$

Which also means that

$\sec 45 = \frac{1}{\frac{\sqrt{2}}{2}}$

which simplifies to

$\sec 45 = \frac{2}{\sqrt{2}}$

The convention is to not have an irrational number in the denominator, so we can multiply the top and bottom by $\sqrt{2}$. We get

$\sec 45 = \frac{2 \sqrt{2}}{\sqrt{2} \sqrt{2}}$

$\sec 45 = \frac{\cancel{2} \sqrt{2}}{\cancel{2}}$

$\sec 45 = \sqrt{2}$

Hope this helps!