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Answers without the blur. Sign up and see all textbooks for free! Q18.

Expert-verified Found in: Page 551 ### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508 # Use an Addition orSubtraction Formula to write the expression as a trigonometricthe function of one number, and then find its exact value. $\frac{\mathrm{tan}\frac{\pi }{18}+\mathrm{tan}\frac{\pi }{9}}{1-\mathrm{tan}\frac{\pi }{18}\mathrm{tan}\frac{\pi }{9}}$

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

See the step by step solution

## Step 1. Given information.

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

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 $\frac{\pi }{18}$ and $\frac{\pi }{9}$ because according to the addition formula for tangents,

$\mathrm{tan}\left(\alpha +\beta \right)=\frac{\mathrm{tan}\alpha +\mathrm{tan}\beta }{1-\mathrm{tan}\alpha \mathrm{tan}\beta }$ where $\alpha =\frac{\pi }{18}$ and $\beta =\frac{\pi }{9}$

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

$\frac{\mathrm{tan}\frac{\pi }{18}+\mathrm{tan}\frac{\pi }{9}}{1-\mathrm{tan}\frac{\pi }{18}\mathrm{tan}\frac{\pi }{9}}=\mathrm{tan}\left(\frac{\pi }{18}+\frac{\pi }{9}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{tan}\frac{\pi }{6}=\frac{\sqrt{3}}{3}$ ### Want to see more solutions like these? 