What is the value of tan18^o

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$\frac{\sqrt{5} - 1}{\sqrt{10 + 2 \sqrt{5}}}$

$\frac{\sqrt{10 + 2 \sqrt{5}}}{\sqrt{5} - 1}$

$\frac{\sqrt{5} + 1}{\sqrt{5} + 1}$

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answer is A.

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Detailed Solution

$\tan 18 ° = \frac{\sin 18 °}{\cos 18 °} = \frac{\frac{\sqrt{5} - 1}{4}}{\frac{\sqrt{10 + 2 \sqrt{5}}}{4}} \tan 18 ° = \frac{\sqrt{5} - 1}{\sqrt{10 + 2 \sqrt{5}}}$

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