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

Question

Aicharya · Accepted Answer

$\begin{aligned} \tan & 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ} \ & =\tan 81^{\circ}+\tan 9^{\circ}-\left(\tan 63^{\circ}+\tan 27^{\circ}\right) \ & =\cot 9^{\circ}+\tan 9^{\circ}-\left(\cot 27^{\circ}+\tan 27^{\circ}\right) \ & =\frac{\cos ^2 9^{\circ}+\sin ^2 9^{\circ}}{\sin 9^{\circ} \cdot \cos 9^{\circ}}-\frac{\cos ^2 27^{\circ}+\sin ^2 27^{\circ}}{\sin 27 \cdot \cos 27} \ & =\frac{1}{\sin 9^{\circ} \cdot \cos 9^{\circ}}-\frac{1}{\sin 27^{\circ} \cdot \cos 27^{\circ}} \ & =\frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}}=2 \frac{\sin 54-\sin 18}{\sin 54 \cdot \sin 18} \ & =\frac{4 \cos 36 \cdot \sin 18}{\cos 36 \cdot \sin 18}=4 \end{aligned}$

\(\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}=\)

Solution