# What is Sec(tan^-1(2))?

Oct 4, 2015

$\sec \left(\arctan \left(2\right)\right) = \sqrt{5}$

#### Explanation:

From the trigonometric identity ${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$, divide both sides by ${\cos}^{2} \left(\theta\right)$

${\sin}^{2} \frac{\theta}{\cos} ^ 2 \left(\theta\right) + {\cos}^{2} \frac{\theta}{\cos} ^ 2 \left(\theta\right) = \frac{1}{\cos} ^ 2 \left(\theta\right)$

${\tan}^{2} \left(\theta\right) + 1 = {\sec}^{2} \left(\theta\right)$

Substitute $\theta$ for $\arctan \left(2\right)$

${\tan}^{2} \left(\arctan \left(2\right)\right) + 1 = {\sec}^{2} \left(\arctan \left(2\right)\right)$

Since $\tan \left(\arctan \left(x\right)\right) = x$ axiomatically, we have that

${\sec}^{2} \left(\arctan \left(2\right)\right) = {\left(2\right)}^{2} + 1$
${\sec}^{2} \left(\arctan \left(2\right)\right) = 5$

Take the root

$\sec \left(\arctan \left(2\right)\right) = \pm \sqrt{5}$

To pick the sign look at the range of the arctangent. During this range $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ the cosine is always positive, and therefore so is the secant

$\sec \left(\arctan \left(2\right)\right) = \sqrt{5}$