Without using tables, show that $tan 9^{o} - tan 27^{o} - tan 63^{o} + tan 81^{o} = 4$

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Solution

$tan 9^{o} - tan 27^{o} - tan 63^{o} + tan 81^{o} = tan 9^{o} - tan 27^{o} - tan (90^{o} - 27^{o}) + tan (90^{o} - 9^{o}) = tan 9^{o} + cot 9^{o} - tan 27^{o} - cot 27^{o} = \frac{sin 9^{o}}{cos 9^{o}} + \frac{cos 9^{o}}{sin 9^{o}} - \frac{sin 27^{o}}{cos 27^{o}} - \frac{cos 27^{o}}{sin 27^{o}} = \frac{1}{sin 9^{o} . c o s 9^{o}} - \frac{1}{sin 27^{o} cos 27^{o}} = \frac{2}{2 sin 9^{o} . c o s 9^{o}} - \frac{2}{2 sin 27^{o} cos 27^{o}} = \frac{2}{sin 18^{o}} - \frac{2}{sin 54^{o}} = \frac{2}{\frac{\sqrt{5} - 1}{4}} - \frac{2}{\frac{\sqrt{5} + 1}{4}} = 8 * [\frac{\sqrt{5} + 1 - \sqrt{5} + 1}{{\sqrt{5}}^{2} - 1}] = 8 * \frac{2}{4} = 4$

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