What is the exact value of $cos ({tan}^{- 1} (\frac{2}{3}))$ $cos(tan^-1 (2/3))$ ?

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Answer 1 · 2015-04-15T14:14:15

${tan}^{- 1} (\frac{2}{3})$ $tan^-1(2/3)$ is a $t$ $t$ in $(- \frac{π}{2}, \frac{π}{2})$ $(- pi/2, pi/2)$ with $tan t = \frac{2}{3}$ $tan t = 2/3$

$t$ $t$ is in quadrant 1 or 2, so $cos t$ $cost$ is positive.

The problem has become:
$tan t = \frac{2}{3}$ $tan t = 2/3$ and $cos t > 0$ $cost > 0$ .Find $cos t$ $cost$ .

Use your favorite method for solving such problems.

It's hard to draw and label a triangle or unit circle here, so I'll use the identities:

$cos t = \frac{1}{sec t}$ $cost = 1/sect$ and ${tan}^{2} t + 1 = {sec}^{2} t$ $tan^2t + 1= sec^2t$

${sec}^{2} t = {(\frac{2}{3})}^{2} + 1 = \frac{4}{9} + 1 = \frac{13}{9}$ $sec^2t = (2/3)^2 + 1 =4/9 + 1 = 13/9$

So, $sec t = \frac{\sqrt{13}}{3}$ $sect = sqrt13/3$ (Remember $cos t > 0$ $cost > 0$ , so $sec t > 0$ $sect > 0$ as well)

$cos t = \frac{3}{\sqrt{13}} = \frac{3 \sqrt{13}}{13}$ $cost = 3/sqrt13 = (3sqrt13)/13$

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