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

Solution:

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