First slide

Compound angles

Question

$\tan \frac{2 π}{5} - \tan \frac{π}{15} - \sqrt{3} \tan \frac{2 π}{5} \tan \frac{π}{15}$ is equal to

Moderate

A

$- \sqrt{3}$
B

$1 / \sqrt{3}$
C

1
D

$\sqrt{3}$

Solution

We have,
$\begin{array}{l} \frac{\tan \frac{6 π}{15} - \tan \frac{π}{15}}{1 + \tan \frac{6 π}{15} \tan \frac{π}{15}} = \tan \frac{π}{3} \\ = \tan \frac{6 π}{15} - \tan \frac{π}{15} = \sqrt{3} + \sqrt{3} \tan \frac{6 π}{15} \tan \frac{π}{15} \\ = t an \frac{6 π}{15} - \tan \frac{π}{15} - \sqrt{3} \tan \frac{6 π}{15} \tan \frac{π}{15} = \sqrt{3} \end{array}$

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