Calculate $\tan9^{\circ}-\tan27^{\circ}-\tan63^{\circ}+\tan81^{\circ}$ Calculate $\tan9^{\circ}-\tan27^{\circ}-\tan63^{\circ}+\tan81^{\circ}$?
The correct answer should be 4. 
 A: You'd better to know
$$\sin 2\theta = 2\sin \theta \cos \theta$$
$$\sin \alpha-\sin \beta=
2\cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}$$
Then we have
\begin{align*}
  \tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ} &=
  \frac{\sin 9^{\circ}}{\cos 9^{\circ}}-
  \frac{\sin 27^{\circ}}{\cos 27^{\circ}}-
  \frac{\cos 27^{\circ}}{\sin 27^{\circ}}+
  \frac{\cos 9^{\circ}}{\sin 9^{\circ}} \\
  &= \frac{1}{\sin 9^{\circ} \cos 9^{\circ}}-
     \frac{1}{\sin 27^{\circ} \cos 27^{\circ}} \\
  &= \frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}} \\
  &= \frac{2(\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{align*}
A: $\tan9^{\circ} + \tan81^{\circ} = \dfrac{\sin9^{\circ}}{\cos9^{\circ}} + \dfrac{\sin81^{\circ}}{\cos81^{\circ}} = \dfrac{\sin(9^{\circ} + 81^{\circ})}{\cos9^{\circ}\cos81^{\circ}} = \dfrac{1}{\cos9^{\circ}\cos81^{\circ}} = \dfrac{1}{\cos9^{\circ}\sin9^{\circ}} = \dfrac{2}{\sin18^{\circ}}$. Similarly:
$\tan27^{\circ} + \tan63^{\circ} = \dfrac{1}{\cos27^{\circ}\cos63^{\circ}} = \dfrac{1}{\cos27^{\circ}\sin27^{\circ}} = \dfrac{2}{\sin54^{\circ}}$.
Now let $x = \sin18^{\circ}$, then from $\cos36^{\circ} = \sin54^{\circ}$ we have:
$1 - 2x^2 = 3x - 4x^3 \to 4x^3 - 2x^2 - 3x + 1 = 0 \to (x-1)(4x^2 + 2x - 1) = 0$, since $x \neq 1$, $x = \dfrac{\sqrt{5} - 1}{4} = \sin18^{\circ}$, and $\sin54^{\circ} = 1 - 2x^2 = 1 - 2\left(\dfrac{\sqrt{5} - 1}{4}\right)^2 = \dfrac{\sqrt{5} + 1}{4}$. Thus:
$S = \dfrac{2}{\dfrac{\sqrt{5} - 1}{4}} -\dfrac{2}{\dfrac{\sqrt{5} + 1}{4}} = 4$
A: Observe that $\displaystyle\tan(5\cdot27^\circ)=\tan(180^\circ-45^\circ)=-\tan45^\circ=-1$
$\displaystyle\implies\tan(-27^\circ\cdot5)=-\tan(5\cdot27^\circ)=1$
Similarly we can show that  $\displaystyle\tan5x=1$ for $\displaystyle x=-27^\circ,-63^\circ,9^\circ,81^\circ,45^\circ$
From  Sum of tangent functions where arguments are in specific arithmetic series, $$\tan5x=\frac{\binom51t-\binom53t^3+t^5}{1-\binom52t^2+\binom54t^4}\text{ where } t=\tan x$$
$$5x=45^\circ\implies t^5-5t^4\tan45^\circ+\cdots=0$$ has the roots $\displaystyle \tan(-27^\circ),\tan(-63^\circ),\tan9^\circ,\tan81^\circ,\tan45^\circ$
Using Vieta's formula,
$\displaystyle\tan(-27^\circ)+\tan(-63^\circ)+\tan9^\circ+\tan81^\circ+\tan45^\circ=\frac51$
See also :  Prove the trigonometric identity $(35)$
