Question

$\tan 9-\tan 27-\tan 63+\tan 81=$

Answer 1

Given

$\tan 81+\tan 9-(\tan 63+\tan 27)$

$=\cot 9+\tan 9-(\cot 27+\tan 27)$

$=\frac{(\cos 9{)}^2+(\sin 9{)}^2}{\sin 9.\cos 9}-\frac{(\cos 27{)}^2+(\sin 27{)}^2}{\sin 27.\cos 27}$

$=\frac{1}{\sin 9.\cos 9}-\frac{1}{\sin 27.\cos 27}$

$=\frac{2}{\sin 18}-\frac{2}{\sin 54}$

$=2.\frac{\sin 54-\sin 18}{\sin 54.\sin 18}$

$=\frac{4\cos 36\sin 18}{\cos 36.\sin 18}$

$=4$