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Q 88.

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

# Proving IdentitiesVerify the identity $\frac{\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{u}\mathbf{-}\mathbf{1}}{\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{u}\mathbf{+}\mathbf{1}}\mathbf{=}\frac{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{u}\mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{u}}{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{u}\mathbf{+}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{u}}$

The expression $\frac{\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{u}\mathbf{-}\mathbf{1}}{\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{u}\mathbf{+}\mathbf{1}}\mathbf{=}\frac{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{u}\mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{u}}{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{u}\mathbf{+}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{u}}$ is an identity.

See the step by step solution

## Step 1. Given information

An expression $\frac{\mathrm{sec}u-1}{\mathrm{sec}u+1}=\frac{\mathrm{tan}u-\mathrm{sin}u}{\mathrm{tan}u+\mathrm{sin}u}$

## Step 2. Concept used

For such a query, divide the expression into two halves, LHS and RHS. Separate them and make them simpler. If two outcomes are identical, it's an identity.

## Step 3. Calculation

Now, simplify LHS of $\frac{\mathrm{sec}u-1}{\mathrm{sec}u+1}=\frac{\mathrm{tan}u-\mathrm{sin}u}{\mathrm{tan}u+\mathrm{sin}u}$

We can use identity as $\mathrm{sec}\left(u\right)=\frac{1}{\mathrm{cos}\left(u\right)}$,

$\begin{array}{l}=\frac{\frac{1}{\mathrm{cos}u}-1}{\frac{1}{\mathrm{cos}u}+1}\\ =\frac{\frac{1-\mathrm{cos}u}{\mathrm{cos}u}}{\frac{1+\mathrm{cos}u}{\mathrm{cos}u}}\\ =\frac{1-\mathrm{cos}u}{\mathrm{cos}u}×\frac{\mathrm{cos}u}{1+\mathrm{cos}u}\\ =\frac{1-\mathrm{cos}u}{1+\mathrm{cos}u}\end{array}$

Multiply and divide by $\mathrm{sin}u$

$\begin{array}{l}=\frac{\mathrm{sin}u}{\mathrm{sin}u}×\frac{1-\mathrm{cos}u}{1+\mathrm{cos}u}\\ =\frac{\mathrm{sin}u-\mathrm{sin}u×\mathrm{cos}u}{\mathrm{sin}u+\mathrm{sin}u×\mathrm{cos}u}\end{array}$

Now multiply and divide by $\frac{1}{\mathrm{cos}u}$

$\begin{array}{l}=\frac{\frac{1}{\mathrm{cos}u}}{\frac{1}{\mathrm{cos}u}}×\frac{\mathrm{sin}u-\mathrm{sin}u×\mathrm{cos}u}{\mathrm{sin}u+\mathrm{sin}u×\mathrm{cos}u}\\ =\frac{\frac{\mathrm{sin}u}{\mathrm{cos}u}-\mathrm{sin}u×\frac{\mathrm{cos}u}{\mathrm{cos}u}}{\frac{\mathrm{sin}u}{\mathrm{cos}u}+\mathrm{sin}u×\frac{\mathrm{cos}u}{\mathrm{cos}u}}\\ \because \mathrm{tan}u=\frac{\mathrm{sin}u}{\mathrm{cos}u}\\ =\frac{\mathrm{tan}u-\mathrm{sin}u}{\mathrm{tan}u+\mathrm{sin}u}\end{array}$

RHS of the equation is $\frac{\mathrm{tan}u-\mathrm{sin}u}{\mathrm{tan}u+\mathrm{sin}u}$

Both RHS and LHS are equal. Hence it is an identity

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