Suggested languages for you:

Americas

Europe

Q69.

Expert-verified
Found in: Page 543

### 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 $\frac{1+\mathrm{tan}x}{1-\mathrm{tan}x}=\frac{\mathrm{cos}x+\mathrm{sin}x}{\mathrm{cos}x-\mathrm{sin}x}$

The expression $\frac{1+\mathrm{tan}x}{1-\mathrm{tan}x}=\frac{\mathrm{cos}x+\mathrm{sin}x}{\mathrm{cos}x-\mathrm{sin}x}$ is an identity.

See the step by step solution

An expression

## Step 2. Concept used.

For proving that the given expression is an identity, divide expression in two parts like LHS and RHS. After completing splitting, simplify both independently. If both results are equal then given expression is an identity.

## Step 3. Calculation.

Now, simplify LHS of $\frac{1+\mathrm{tan}x}{1-\mathrm{tan}x}=\frac{\mathrm{cos}x+\mathrm{sin}x}{\mathrm{cos}x-\mathrm{sin}x}$:

Use substitution as $\mathrm{tan}x=\frac{\mathrm{sin}x}{\mathrm{cos}x}$,

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

RHS of the equation is $\frac{\mathrm{cos}x+\mathrm{sin}x}{\mathrm{cos}x-\mathrm{sin}x}$

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