Question

In Problems 9–36, find the exact value of each expression.

$\csc \left({\tan }^{-1}1\right)$

Short answer

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

Step-by-step solution

Step 1. Given information.

The given expression is:

$\csc \left({\tan }^{-1}1\right)$

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

$\tan \theta =1$

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

Step 2. Find the value of θ.

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

By using the unit circle, we can say that $\theta$must be equal to $\frac{\mathrm{\pi }}{4}$, so that we can get 1.

$$\begin{gathered} \tan \theta =1 \\ {\tan }^{-1}=\frac{\mathrm{\pi }}{4} \end{gathered}$$

Step 3. Find the exact value of the expression.

$$\begin{gathered} \csc \left({\tan }^{-1}1\right)=\csc \left(\frac{\mathrm{\pi }}{4}\right) \\ \csc \left(\theta \right)=\frac{1}{\sin \theta } \\ \csc \left(\frac{\mathrm{\pi }}{4}\right)=\frac{1}{\sin \left(\frac{\mathrm{\pi }}{4}\right)} \end{gathered}$$

By using the unit circle, we know that $\sin \left(\frac{\mathrm{\pi }}{4}\right)=\left(\frac{\sqrt{2}}{2}\right)$, then

$$\begin{gathered} \cos \left(\frac{\mathrm{\pi }}{4}\right)=\frac{1}{\frac{\sqrt{2}}{2}} \\ =\frac{2}{\sqrt{2}} \\ =\left(\frac{2}{\sqrt{2}}\right)\left(\frac{\sqrt{2}}{\sqrt{2}}\right) \\ =\frac{2\sqrt{2}}{2} \\ =\sqrt{2} \end{gathered}$$

Therefore,$\csc \left({\tan }^{-1}1\right)=\sqrt{2}$.