Q. 105 (a) Show that聽tantan-11+tan-12+... [FREE SOLUTION]

91影视

Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 105 (page 487)

(a) Show that $\tan (\tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3) = 0$
(b) Conclude from part (a) that
$\tan (\tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3) = 0$

Short Answer

Expert verified

(a)The Equation $\tan (\tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3) = 0$ can be obtained by using the sum formula for tangent function as

$\tan (\tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3) = \frac{\tan (\tan^{- 1} 1 + \tan^{- 1} 2) + 3}{1 - \tan (\tan^{- 1} 1 + \tan^{- 1} 2) 路 3} = \frac{\frac{\tan (\tan^{- 1} 1) + \tan (\tan^{- 1} 2)}{1 - \tan (\tan^{- 1} 1) 路 \tan (\tan^{- 1} 2)} + 3}{1 - \frac{\tan (\tan^{- 1} 1) + \tan (\tan^{- 1} 2)}{1 - \tan (\tan^{- 1} 1) 路 \tan (\tan^{- 1} 2)} 路 3} = 0$

(b) the equation $\tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3 = 蟺$ is obtained as

$\tan (\tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3) = 0 \tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3 = \tan^{- 1} 0 \tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3 = 蟺$

Step by step solution

Step 1. Given data

Equations that need to be proven are
$\tan (\tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3) = 0 \tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3 = 蟺$

Step 2. Part (a)

Use Sum formula for the tangent function

$\tan (\tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3) = \tan ((\tan^{- 1} 1 + \tan^{- 1} 2) + \tan^{- 1} 3) = \frac{\tan (\tan^{- 1} 1 + \tan^{- 1} 2) + \tan (\tan^{- 1} 3)}{1 - \tan (\tan^{- 1} 1 + \tan^{- 1} 2) 路 \tan (\tan^{- 1} 3)} = \frac{\tan (\tan^{- 1} 1 + \tan^{- 1} 2) + 3}{1 - \tan (\tan^{- 1} 1 + \tan^{- 1} 2) 路 3}$

Step 3. Apply Sum formula for the tangent function

Use Sum formula for the tangent function

so $\tan (\tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3) = 0$

Step 4. Part (b)

Change inverse trigonometric function into a trigonometric function

$\tan (\tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3) = 0 \tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3 = \tan^{- 1} 0$

for interval $[0, 蟺]$

$\tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3 = \tan^{- 1} 0 \tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3 = 蟺$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

(a) Show that $\tan (\tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3) = 0$
(b) Conclude from part (a) that
$\tan (\tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3) = 0$

Short Answer

Step by step solution

Step 1. Given data

Step 2. Part (a)

Step 3. Apply Sum formula for the tangent function

Step 4. Part (b)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Mechanics Maths

Probability and Statistics

Logic and Functions

Decision Maths

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

91影视

(a) Show thattantan-11+tan-12+tan-13=0(b) Conclude from part (a) thattantan-11+tan-12+tan-13=0

Short Answer

Step by step solution

Step 1. Given data

Step 2. Part (a)

Step 3. Apply Sum formula for the tangent function

Step 4. Part (b)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Mechanics Maths

Probability and Statistics

Logic and Functions

Decision Maths

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

(a) Show that $\tan (\tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3) = 0$
(b) Conclude from part (a) that
$\tan (\tan^{- 1} 1 + \tan^{- 1} 2 + \tan^{- 1} 3) = 0$