How do you evaluate $sec (arctan 2)$ $sec(arctan 2)$ ?

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Answer 1 · 2016-05-15T02:12:26

$\pm \sqrt{5}$ $+-sqrt 5$

Let a = arc tan 2. Then, tan a = 2 > 0. a is in either the 1st quadrant or the 3rd.

So, the given expression $sec a = \pm \sqrt{5}$ $sec a =+-sqrt 5$ ,

using $sec a = \pm \sqrt{1 + {tan}^{2} a}$ $sec a =+-sqrt(1+tan^2 a)$ ..

How do you evaluate sec(arctan2)sec(arctan 2)?