# How would you solve: tan(15 degrees)/1-tan^2(15 degrees)? (show steps) Thanks

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Mar 9, 2018

$\tan {15}^{\circ} / \left(1 - {\tan}^{2} {15}^{\circ}\right) = \frac{1}{2 \sqrt{3}}$

#### Explanation:

We use the identity $\frac{2 \tan A}{1 - {\tan}^{2} A} = \tan 2 A$

Now if $A = {15}^{\circ}$

$\frac{2 \tan {15}^{\circ}}{1 - {\tan}^{2} {15}^{\circ}} = \tan \left(2 \times {15}^{\circ}\right)$

or $\frac{2 \tan {15}^{\circ}}{1 - {\tan}^{2} {15}^{\circ}} = \tan {30}^{\circ} = \frac{1}{\sqrt{3}}$

Hence $\tan {15}^{\circ} / \left(1 - {\tan}^{2} {15}^{\circ}\right) = \frac{1}{2 \sqrt{3}}$

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