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Q69.

Expert-verified

Precalculus Mathematics for Calculus

Found in: Page 543

Book edition 7th Edition

Author(s) James Stewart, Lothar Redlin, Saleem Watson

Pages 948 pages

ISBN 9781337067508

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Short Answer

Proving Identities
Verify the identity $\frac{1 + \tan x}{1 - \tan x} = \frac{\cos x + \sin x}{\cos x - \sin x}$

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

An expression $\frac{1 + \tan x}{1 - \tan x} = \frac{\cos x + \sin x}{\cos x - \sin x}$

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

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

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

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

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

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