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Answers without the blur. Sign up and see all textbooks for free! Q 16.

Expert-verified Found in: Page 457 ### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465 # In Problems 9–36, find the exact value of each expression.$\mathrm{sec}\left({\mathrm{tan}}^{-1}\sqrt{3}\right)$

The value of the expression $\mathrm{sec}\left({\mathrm{tan}}^{-1}\sqrt{3}\right)=2$.

See the step by step solution

## Step 1. Given information.

The given expression is:

$\mathrm{sec}\left({\mathrm{tan}}^{-1}\sqrt{3}\right)$

Let's take $\theta ={\mathrm{tan}}^{-1}\sqrt{3}$,

$\mathrm{tan}\theta =\sqrt{3}$

$-\frac{\mathrm{\pi }}{2}\le \theta \le \frac{\mathrm{\pi }}{2}$ is the domain of the function.

## Step 2. Find the value of θ.

Initially, we have function ${\mathrm{tan}}^{-1}$ and the bounds of tan function are going to be in the interval $\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$. It represents the range of ${\mathrm{tan}}^{-1}$.

By using the unit circle, we can say that $\theta$ must be equal to $\frac{\mathrm{\pi }}{3}$, so that we can get $\sqrt{3}$.

$\mathrm{tan}\theta =\sqrt{3}\phantom{\rule{0ex}{0ex}}{\mathrm{tan}}^{-1}\sqrt{3}=\frac{\mathrm{\pi }}{3}$

## Step 3. Find the exact value of the expression.

$\mathrm{sec}\left({\mathrm{tan}}^{-1}\sqrt{3}\right)=\mathrm{sec}\left(\frac{\mathrm{\pi }}{3}\right)\phantom{\rule{0ex}{0ex}}\mathrm{sec}\left(\theta \right)=\frac{1}{\mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}\mathrm{sec}\left(\frac{\mathrm{\pi }}{3}\right)=\frac{1}{\mathrm{cos}\left(\frac{\mathrm{\pi }}{3}\right)}$

By using the unit circle, we know that $\mathrm{cos}\left(\frac{\mathrm{\pi }}{3}\right)=\frac{1}{2}$, then

Therefore, $\mathrm{sec}\left({\mathrm{tan}}^{-1}\sqrt{3}\right)=2$. ### Want to see more solutions like these? 