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Q 96.

Expert-verified
Found in: Page 544

### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508

# Determining Identities GraphicallyGraph $\mathbit{f}$ and $\mathbit{g}$ in the same viewing rectangle. Do the graphs suggest that the equation $\mathbit{f}\mathbf{\left(}\mathbit{x}\mathbf{\right)}\mathbf{=}\mathbit{g}\mathbf{\left(}\mathbit{x}\mathbf{\right)}$ is an identity? Prove your answer$\mathbit{f}\mathbf{\left(}\mathbit{x}\mathbf{\right)}\mathbf{=}\mathbit{t}\mathbit{a}\mathbit{n}\mathbit{x}\mathbf{\left(}\mathbf{1}\mathbf{+}\mathbit{s}\mathbit{i}\mathbit{n}\mathbit{x}\mathbf{\right)}\mathbf{,}\mathbf{\text{ }}\mathbit{g}\mathbf{\left(}\mathbit{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{x}}{\mathbf{1}\mathbf{+}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}}$

The expression $\mathbit{f}\mathbf{\left(}\mathbit{x}\mathbf{\right)}\mathbf{=}\mathbit{g}\mathbf{\left(}\mathbit{x}\mathbf{\right)}$ is an not identity.

See the step by step solution

## Step 1. Given information

Two functions as-

$\begin{array}{l}f\left(x\right)=\mathrm{tan}x\left(1+\mathrm{sin}x\right)\\ \text{ }g\left(x\right)=\frac{\mathrm{sin}x\mathrm{cos}x}{1+\mathrm{sin}x}\end{array}$

## Step 2. Concept used

Create distinct graphs for both functions on the same graph. Both functions are equivalent if both graphs coincide with one other. As a result, combining them into a single equation yields an identity.

## Step 3. Calculation

Now, draw graphs of both functions:

Since both graphs are not coincide each other. Therefore $f\left(x\right)=g\left(x\right)$ is not an identity