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Q18.

Expert-verified

Precalculus Mathematics for Calculus

Found in: Page 551

Book edition 7th Edition

Author(s) James Stewart, Lothar Redlin, Saleem Watson

Pages 948 pages

ISBN 9781337067508

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Short Answer

Use an Addition or
Subtraction Formula to write the expression as a trigonometric
the function of one number, and then find its exact value.
$\frac{\tan \frac{π}{18} + \tan \frac{π}{9}}{1 - \tan \frac{π}{18} \tan \frac{π}{9}}$

The exact value of this equation is $\frac{\sqrt{3}}{3}$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

The given function is $\frac{\tan \frac{π}{18} + \tan \frac{π}{9}}{1 - \tan \frac{π}{18} \tan \frac{π}{9}}$

Use an addition or subtraction Formula to write the expression.

Then appraise the expression to obtain the final answer.

Step 2. Determining the value.

The expression can be expressed as the tangent of the sum of the angles $\frac{π}{18}$ and $\frac{π}{9}$ because according to the addition formula for tangents,

$\tan (α + β) = \frac{\tan α + \tan β}{1 - \tan α \tan β}$ where $α = \frac{π}{18}$ and $β = \frac{π}{9}$

Now, $\tan \frac{π}{6} = \frac{\sqrt{3}}{3}$

$\frac{\tan \frac{π}{18} + \tan \frac{π}{9}}{1 - \tan \frac{π}{18} \tan \frac{π}{9}} = \tan (\frac{π}{18} + \frac{π}{9}) = \tan \frac{π}{6} = \frac{\sqrt{3}}{3}$

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