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

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Answer 1 · 2015-10-04T01:40:23

$sec(arctan(2)) = sqrt(5)$

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

$sin^2(theta)/cos^2(theta) + cos^2(theta)/cos^2(theta) = 1/cos^2(theta)$

$tan^2(theta) + 1 = sec^2(theta)$

Substitute $theta$ for $arctan(2)$

$tan^2(arctan(2)) + 1 = sec^2(arctan(2))$

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

$sec^2(arctan(2)) = (2)^2+1$
$sec^2(arctan(2)) = 5$

Take the root

$sec(arctan(2)) = +-sqrt(5)$

To pick the sign look at the range of the arctangent. During this range $(-pi/2,pi/2)$ the cosine is always positive, and therefore so is the secant

$sec(arctan(2)) = sqrt(5)$