tan⁡81∘−tan⁡63∘−tan⁡27∘+tan⁡9∘ equals

tan⁡81∘−tan⁡63∘−tan⁡27∘+tan⁡9∘=tan⁡81∘+tan⁡9∘−tan⁡63∘+tan⁡27∘=sin⁡81∘cos⁡81∘+sin⁡9∘cos⁡9∘−sin⁡63∘cos⁡63∘+sin⁡27∘cos⁡27∘=sin⁡81∘cos⁡9∘+sin⁡9∘cos⁡81∘cos⁡9∘cos⁡81∘−sin⁡63∘cos⁡27∘+sin⁡27∘cos⁡63∘cos⁡27∘cos⁡63∘=sin⁡81∘+9∘cos⁡9∘cos⁡90∘−9∘−sin⁡63∘+27∘cos⁡27∘cos⁡90∘−27∘=sin⁡90∘cos⁡9∘sin⁡9∘−sin⁡90∘cos⁡27∘sin⁡27∘=1sin⁡9∘cos⁡9∘×22−1sin⁡27∘cos⁡27∘×22=2sin⁡18∘−2sin⁡54∘=2sin⁡54∘−sin⁡18∘sin⁡18∘sin⁡54∘=22cos⁡36∘sin⁡18∘sin⁡18∘cos⁡36∘=4

tan⁡81∘−tan⁡63∘−tan⁡27∘+tan⁡9∘ equals

tan⁡81∘−tan⁡63∘−tan⁡27∘+tan⁡9∘=tan⁡81∘+tan⁡9∘−tan⁡63∘+tan⁡27∘=sin⁡81∘cos⁡81∘+sin⁡9∘cos⁡9∘−sin⁡63∘cos⁡63∘+sin⁡27∘cos⁡27∘=sin⁡81∘cos⁡9∘+sin⁡9∘cos⁡81∘cos⁡9∘cos⁡81∘−sin⁡63∘cos⁡27∘+sin⁡27∘cos⁡63∘cos⁡27∘cos⁡63∘=sin⁡81∘+9∘cos⁡9∘cos⁡90∘−9∘−sin⁡63∘+27∘cos⁡27∘cos⁡90∘−27∘=sin⁡90∘cos⁡9∘sin⁡9∘−sin⁡90∘cos⁡27∘sin⁡27∘=1sin⁡9∘cos⁡9∘×22−1sin⁡27∘cos⁡27∘×22=2sin⁡18∘−2sin⁡54∘=2sin⁡54∘−sin⁡18∘sin⁡18∘sin⁡54∘=22cos⁡36∘sin⁡18∘sin⁡18∘cos⁡36∘=4

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$\tan 81^{\circ} - \tan 63^{\circ} - \tan 27^{\circ} + \tan 9^{\circ}$ equals

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Detailed Solution

$\begin{array}{l} \tan 81^{\circ} - \tan 63^{\circ} - \tan 27^{\circ} + \tan 9^{\circ} = [\tan 81^{\circ} + \tan 9^{\circ}) - [\tan (63^{\circ}) + \tan (27^{\circ})] \\ = [\frac{\sin 81^{\circ}}{\cos 81^{\circ}} + \frac{\sin 9^{\circ}}{\cos 9^{\circ}}] - [\frac{\sin 63^{\circ}}{\cos 63^{\circ}} + \frac{\sin 27^{\circ}}{\cos 27^{\circ}}] \\ = [\frac{\sin 81^{\circ} \cos 9^{\circ} + \sin 9^{\circ} \cos 81^{\circ}}{\cos 9^{\circ} \cos 81^{\circ}}] - [\frac{\sin 63^{\circ} \cos 27^{\circ} + \sin 27^{\circ} \cos 63^{\circ}}{\cos 27^{\circ} \cos 63^{\circ}}] \\ = \frac{[\sin (81^{\circ} + 9^{\circ})]}{\cos 9^{\circ} \cos (90^{\circ} - 9^{\circ})} - \frac{[\sin (63^{\circ} + 27^{\circ})]}{\cos 27^{\circ} \cos (90^{\circ} - 27^{\circ})} \\ = \frac{\sin 90^{\circ}}{\cos 9^{\circ} \sin 9^{\circ}} - \frac{\sin 90^{\circ}}{\cos 27^{\circ} \sin 27^{\circ}} \\ = \frac{1}{\sin 9^{\circ} \cos 9^{\circ}} \times \frac{2}{2} - \frac{1}{\sin 27^{\circ} \cos 27^{\circ}} \times \frac{2}{2} \\ = \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}} \\ = 2 [\frac{\sin 54^{\circ} - \sin 18^{\circ}}{\sin 18^{\circ} \sin 54^{\circ}}] = 2 [\frac{2 \cos 36^{\circ} \sin 18^{\circ}}{\sin 18^{\circ} \cos 36^{\circ}}] = 4 \end{array}$