Chapter 6.7, Problem 48E

To determine

To find: the tangent function that has the same graphs as the given function.

Answer to Problem 48E

  $\begin{array}{l}y=\tan \left(\frac{\pi }{2}-\theta \right)\\ y=-\tan \left(\frac{\pi }{2}+\theta \right)\end{array}$

Explanation of Solution

Given information:

  $y=\cot \theta$

Calculation:

Let $x=\theta$ . The graph of cotangent and tangent crosses $x$ -axis at $\theta =\frac{\pi }{2}+n\pi$ and $\theta =n\pi$ respectively, where $n$ is any integer.

If the tangent function is either shifted horizontally by $\frac{\pi }{2}$ or shifting horizontally by $\frac{\pi }{2}$ and reflecting the same on vertical reflection, then the functions can have same graph as that of given function.

The functions can be evaluated as follows:

  $\begin{array}{c}y=\tan \left(-\left(\theta -\frac{\pi }{2}\right)\right)\\ =\tan \left(-\theta +\frac{\pi }{2}\right)\\ =\tan \left(\frac{\pi }{2}-\theta \right)\end{array}$

  $\begin{array}{c}y=-\tan \left(\theta -\frac{\pi }{2}\right)\\ =-\tan \left(\frac{\pi }{2}+\theta \right)\end{array}$

Thus, the tangent function that has the same graphs as $y=\cot \theta$ is $y=\tan \left(\frac{\pi }{2}-\theta \right)$ and $y=-\tan \left(\frac{\pi }{2}+\theta \right)$ .