# 005 Sample Final A, Question 15

Question Find an equivalent algebraic expression for the following,

${\displaystyle \cos(\tan ^{-1}(x))}$
Foundations
1) ${\displaystyle \tan ^{-1}(x)}$ can be thought of as ${\displaystyle \tan ^{-1}\left({\frac {x}{1}}\right),}$ and this now refers to an angle in a triangle. What are the side lengths of this triangle?
1) The side lengths are 1, x, and ${\displaystyle {\sqrt {1+x^{2}}}.}$
First, let ${\displaystyle \theta =\tan ^{-1}(x)}$. Then, ${\displaystyle \tan(\theta )=x}$.
Now, we draw the right triangle corresponding to ${\displaystyle \theta }$. Two of the side lengths are 1 and x and the hypotenuse has length ${\displaystyle {\sqrt {x^{2}+1}}}$.
Since ${\displaystyle \cos(\theta )={\frac {\mathrm {opposite} }{\mathrm {hypotenuse} }}}$, ${\displaystyle \cos(\tan ^{-1}(x))=\cos(\theta )={\frac {1}{\sqrt {x^{2}+1}}}}$.
${\displaystyle {\frac {1}{\sqrt {x^{2}+1}}}}$