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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 37 (page 474)

In problems 19-98, establish each identity.
$\sec (u) - \tan (u) = \frac{\cos (u)}{1 + \sin (u)}$

Short Answer

Expert verified

We have proved the identity $\sec (u) - \tan (u) = \frac{\cos (u)}{1 + \sin (u)}$ .

Step by step solution

01

Step 1. Given information.

We have:

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

02

Step 2. Establish the identity.

Take the left hand side:

$\sec (u) - \tan (u)$

Apply property: $\sec (u) = \frac{1}{\cos (u)}, \tan (u) = \frac{\sin (u)}{\cos (u)}$

$= \frac{1}{\cos (u)} - \frac{\sin (u)}{\cos (u)} = \frac{1 - \sin (u)}{\cos (u)}$

Multiply numerator and denominator with $1 + \sin (u)$ :

$= \frac{1 - \sin (u)}{\cos (u)} 脳 \frac{1 + \sin (u)}{1 + \sin (u)} = \frac{(1 - \sin (u)) (1 + \sin (u))}{\cos (u) (1 + \sin (u))} = \frac{1^{2} - \sin^{2} u}{\cos (u) (1 + \sin (u))} = \frac{\cos^{2} (u)}{\cos (u) (1 + \sin (u))} = \frac{\cos (u)}{1 + \sin (u)}$

Hence, proved.

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Most popular questions from this chapter

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