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Q 88.

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Precalculus Mathematics for Calculus

Found in: Page 544

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 $\frac{s e c u - 1}{s e c u + 1} = \frac{t a n u - s i n u}{t a n u + s i n u}$

The expression $\frac{s e c u - 1}{s e c u + 1} = \frac{t a n u - s i n u}{t a n u + s i n u}$ 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 $\frac{\sec u - 1}{\sec u + 1} = \frac{\tan u - \sin u}{\tan u + \sin u}$

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

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

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

Multiply and divide by $\sin u$

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

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

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

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

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

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