Q. 53 - Precalculus Enhanced with Graphing Utilities

Question

Graph$y=\tan x and y=-cot\left(x+\frac{\mathrm{\pi }}{2}\right)$

Do you think that $\tan x = y=-cot\left(x+\frac{\mathrm{\pi }}{2}\right)$?

Step-by-Step Solution

Answer

Graph of$y=\tan x \& y=-cot\left(x+\frac{\mathrm{\pi }}{2}\right)$are

Graph of both functions are same so$\tan x=-cot\left(x+\frac{\mathrm{\pi }}{2}\right)$

Step 1. Given data

The given functions are

$$\begin{gathered} y=\tan x \\ y=-cot\left(x+\frac{\mathrm{\pi }}{2}\right) \end{gathered}$$

Step 2. Points coordinates of the tangent function

Consider different values of x and determine coordinates of several points of $y=\tan x$

Step 3. Graph of the tangent function

Plot the graph of $y=\tan x$ with the help of points

Step 4. Points coordinates of the cotangent function

Consider different values of x and determine coordinates of several points of $y=-cot\left(x+\frac{\mathrm{\pi }}{2}\right)$

Step 5. Graph of the cotangent function

Plot the graph of $y=-cot\left(x+\frac{\mathrm{\pi }}{2}\right)$with the help of points 

Both graph are same so $\tan x=-cot\left(x+\frac{\mathrm{\pi }}{2}\right)$