# How do you simplify Tan(sec^-1(u))?

##### 2 Answers
Apr 27, 2018

$\frac{y}{x}$

#### Explanation:

Trigonometric functions of an angle can be thought of as ratios of the sides of a triangle and its inverse functions are functions of a ratio of that triangle that return an angle.

Let there be a right triangle with vertical leg $y$, horizontal leg $x$ and hypotenuse $r$.
Let an angle $\theta$ be defined as the angle formed from the intersection of $x$ and $r$ measured counter-clockwise.

$\sec \left(\theta\right) = \frac{1}{\cos} \left(\theta\right) = \frac{r}{x}$
so ${\sec}^{-} 1 \left(u\right) = {\sec}^{-} 1 \left(\frac{r}{x}\right) = \theta$
The problem now becomes $\tan \left(\theta\right)$.

We know that $\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right) = \frac{\frac{y}{\cancel{r}}}{\frac{x}{\cancel{r}}}$ = $\frac{y}{x}$

Apr 27, 2018

see below

#### Explanation:

Let ${\sec}^{-} 1 \left(u\right)$ be equal to some $\theta$.
Thus,
$\sec \theta = u$
Squaring both sides,we get,
${\sec}^{2} \theta = {u}^{2}$
${\tan}^{2} \theta + 1$=${u}^{2}$ [we know,${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$]
Thus,
$\tan \theta = \pm \sqrt{{u}^{2} - 1}$

Getting back to the original question,we have,
$\tan \left({\sec}^{-} 1 \left(u\right)\right)$ which on simplification gives,
$\tan \theta$ which is equal to $\pm \sqrt{{u}^{2} - 1}$