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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. 38 (page 504)

In Problem 28-44 Establish each identity
$\frac{\cos (伪 - 尾)}{\cos 伪 \cos 尾} = 1 + \tan 伪 \tan 尾$

Short Answer

Expert verified

LHS=RHS

Step by step solution

01

Step 1. Given information is

$\frac{\cos (伪 - 尾)}{\cos 伪 \cos 尾} = 1 + \tan 伪 \tan 尾$

02

Step 2. using the identitycosα-β=cosαcosβ+sinαsinβ

$\frac{\cos (伪 - 尾)}{\cos 伪 \cos 尾} = \frac{\cos 伪 \cos 尾 + \sin 伪 \sin 尾}{\cos 伪 \cos 尾} \frac{\cos (伪 - 尾)}{\cos 伪 \cos 尾} = \frac{\cos 伪 \cos 尾}{\cos 伪 \cos 尾} + \frac{\sin 伪 \sin 尾}{\cos 伪 \cos 尾} \frac{\cos (伪 - 尾)}{\cos 伪 \cos 尾} = 1 + \tan 伪 \tan 尾$

LHS=RHS

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Most popular questions from this chapter

If $\tan 伪 = x + 1$ and role="math" localid="1646425111199" $\tan 尾 = x - 1,$ show that

role="math" localid="1646426027733" $2 c o t (伪 - b) = x^{2}$

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$\frac{1}{2} \sin^{2} x + C = - \frac{1}{4} \cos (2 x)$

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