Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 104 (page 475)

Show $\sqrt{9 s e c^{2} 胃 - 9} = 3 \tan 胃 i f 蟺鈮� 胃 < \frac{3 蟺}{2}$

Short Answer

Expert verified

It is proved that $\sqrt{9 s e c^{2} 胃 - 9} = 3 \tan 胃$

Step by step solution

Step 1. Given Information

The given equation and interval is $\sqrt{9 s e c^{2} 胃 - 9} = 3 \tan 胃, 蟺鈮� 胃 < \frac{3 蟺}{2}$

Step 2. Calculation

Consider the left hand side of the equation and prove it to the right side of the given equation.

$\sqrt{9 s e c^{2} 胃 - 9} = \sqrt{9 (s e c^{2} 胃 - 1)} = 3 \sqrt{s e c^{2} 胃 - 1} = 3 \sqrt{\frac{1}{\cos^{2} 胃} - 1} = 3 \sqrt{\frac{1 - \cos^{2} 胃}{\cos^{2} 胃}} = 3 \sqrt{\frac{\sin^{2} 胃}{\cos^{2} 胃}} = 3 \sqrt{\tan^{2} 胃} = 3 \tan 胃$

Hence, Proved

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91影视

Show $\sqrt{9 s e c^{2} 胃 - 9} = 3 \tan 胃 i f 蟺鈮� 胃 < \frac{3 蟺}{2}$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Calculation

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91影视

Show9sec2胃-9=3tan胃if蟺鈮�胃<3蟺2

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Calculation

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Most popular questions from this chapter

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Study anywhere. Anytime. Across all devices.

Show $\sqrt{9 s e c^{2} 胃 - 9} = 3 \tan 胃 i f 蟺鈮� 胃 < \frac{3 蟺}{2}$