Chapter 7, Problem 5CT

To determine

Findthe value of $\cot \left({\csc }^{-1}\sqrt{10}\right)$ . Express angles in radians.

Answer to Problem 5CT

  $\boxed{3}$

Explanation of Solution

Given information:

Find the value of $\cot \left({\csc }^{-1}\sqrt{10}\right)$ Express angles in radians.

Calculation:

Let $\theta =\cot \left({\csc }^{-1}\sqrt{10}\right)$

Rewrite the inverse function as $\csc \theta =\sqrt{10}$

  $\csc \theta$ is equivalent to $\frac{1}{\sin \theta }$

  $\csc \theta =\frac{1}{\sin \theta }$

Thus we have $\frac{1}{\sin \theta }=\sqrt{10}$ or $\sin \theta =\frac{1}{\sqrt{10}}$ its equivalent to $\frac{\sqrt{10}}{10}$

Since $\sin \theta =\frac{\sqrt{10}}{10}$ , we have the opposite side of the triangle as $\sqrt{10}$ and the hypotenuse as 10.

Use the Pythagorean Thoerem

Adjacent side $=\sqrt{{10}^2-{\left(\sqrt{10}\right)}^2}$

  $\begin{array}{l}=\sqrt{100-10}\\ =\sqrt{90}\end{array}$

  $\sqrt{90}$ is equivalent to $\sqrt{9\cdot 10}$ or $3\sqrt{10}$

So, $\cos \theta =\frac{3\sqrt{10}}{10}$

Use the identity $\cot \theta =\frac{\cos \theta }{\sin \theta }$

Find the value of $\cot \theta$

  $\cot \theta =\frac{\frac{3\sqrt{10}}{10}}{\frac{\sqrt{10}}{10}}$

  $\begin{array}{l}=\frac{3\sqrt{10}}{10}\cdot \frac{10}{\sqrt{10}}\\ =3\end{array}$

Hence the value of expression is $\boxed{3}$ .