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Q18.

Expert-verifiedFound in: Page 551

Book edition
7th Edition

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

Pages
948 pages

ISBN
9781337067508

**Use an Addition or**

**Subtraction Formula to write the expression as a trigonometric**

**the function of one number, and then find its exact value.**

$\frac{\mathrm{tan}\frac{\pi}{18}+\mathrm{tan}\frac{\pi}{9}}{1-\mathrm{tan}\frac{\pi}{18}\mathrm{tan}\frac{\pi}{9}}$

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

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.

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,

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

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

$\frac{\mathrm{tan}\frac{\pi}{18}+\mathrm{tan}\frac{\pi}{9}}{1-\mathrm{tan}\frac{\pi}{18}\mathrm{tan}\frac{\pi}{9}}=\mathrm{tan}\left(\frac{\pi}{18}+\frac{\pi}{9}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{tan}\frac{\pi}{6}=\frac{\sqrt{3}}{3}$

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