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Question: \[\tan 9{^\circ} - \tan 27{^\circ} - \tan 63{^\circ} + \tan 81{^\circ} =\]...

tan⁡9∘−tan⁡27∘−tan⁡63∘+tan⁡81∘=\tan 9{^\circ} - \tan 27{^\circ} - \tan 63{^\circ} + \tan 81{^\circ} =

A

1/2

B

2

C

4

D

8

Answer

4

Explanation

Solution

tan⁡9o−tan⁡27o−tan⁡63o+tan⁡81o\tan{}9^{o} - \tan{}27^{o} - \tan{}63^{o} + \tan{}81^{o}

=tan⁡9o−tan⁡27o−cot⁡27o+cot⁡9o= \tan{}9^{o} - \tan{}27^{o} - \cot{}27^{o} + \cot{}9^{o}

=(tan⁡9o+cot⁡9o)−(tan⁡27o+cot⁡27o)= (\tan{}9^{o} + \cot{}9^{o}) - (\tan{}27^{o} + \cot{}27^{o})

=cos⁡(9o−9o)sin⁡9ocos⁡9o−cos⁡(27o−27o)sin⁡27o.cos⁡27o=2sin⁡18o−2sin⁡54o= \frac{\cos(9^{o} - 9^{o})}{\sin 9^{o}\cos 9^{o}} - \frac{\cos(27^{o} - 27^{o})}{\sin 27^{o}.\cos 27^{o}} = \frac{2}{\sin 18^{o}} - \frac{2}{\sin 54^{o}}

=2{sin⁡54o−sin⁡18osin⁡18osin⁡54o}=2.2.cos⁡36o.sin⁡18osin⁡18o.sin⁡54o=4= 2\left\{ \frac{\sin 54^{o} - \sin 18^{o}}{\sin 18^{o}\sin 54^{o}} \right\} = 2.\frac{2.\cos 36^{o}.\sin 18^{o}}{\sin 18^{o}.\sin 54^{o}} = 4