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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. 33 (page 503)

In Problem 28-44 Establish each identity
$\frac{c s c 胃}{1 + c s c 胃} = \frac{1 - \sin 胃}{\cos^{2} 胃}$

Short Answer

Expert verified

LHS=RHS

Step by step solution

01

Step 1. Given information is

$\frac{c s c 胃}{1 + c s c 胃} = \frac{1 - \sin 胃}{\cos^{2} 胃}$

02

Step 2. We know that cscθ=1sinθ

$\frac{c s c 胃}{1 + c s c 胃} = \frac{\frac{1}{\sin 胃}}{1 + \frac{1}{\sin 胃}} = \frac{\frac{1}{\sin 胃}}{\frac{\sin 胃 + 1}{\sin 胃}} = \frac{1}{1 + \sin 胃}$

03

Step 3. Step 3. Dividing numerator and denominator by 1-sinθ

$= \frac{1}{1 + \sin 胃} 脳 \frac{1 - \sin 胃}{1 - \sin 胃} = \frac{1 - \sin 胃}{1 - \sin^{2} 胃} = \frac{1 - \sin 胃}{\cos^{2} 胃}$

LHS=RHS

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Most popular questions from this chapter

Rewrite in terms of sine and cosine functions: $c o t (胃) s e c (胃)$

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Show that $\tan^{- 1} v + c o t^{- 1} v = \frac{蟺}{2}$

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