Trigonometry

How many times will this question be repeated on brilliant?

$(\tan 81^\circ+\tan9^\circ)-(\tan 63^\circ+\tan27^\circ)$

$=\left(\dfrac{\overbrace{\sin 90^{\circ}}^{\color{#D61F06}{1}}}{\underbrace{\cos 9^{\circ}\underbrace{\cos 81^{\circ}}_{\sin9^{\circ}}}_{\color{#D61F06}{\dfrac{\sin 18^{\circ}}{2}}}}\right)-\left( \dfrac{\overbrace{\sin 90^{\circ}}^{\color{#D61F06}{1}}}{\underbrace{\cos 27^{\circ}\underbrace{\cos 63^{\circ}}_{\sin 27^{\circ}}}_{\color{#D61F06}{\dfrac{\sin 54^{\circ}}2}}}\right)$

$\large=2\left(\dfrac{1}{\frac{\sqrt 5-1}{4}}-\dfrac{1}{\frac{\sqrt 5+1}{4}}\right)$

$\huge =2\times 2=\boxed{\color{#3D99F6}{4}}$

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In second line I used : $\small{\color{teal}{\tan A+\tan B=\dfrac{\sin (A+B)}{\cos A\cos B}~~,\sin A\cos A=\dfrac{\sin 2A}2}}$

Also, $\small{\sin 18^{\circ}=\dfrac{\sqrt 5-1}{4},\sin 54^{\circ}=\dfrac{\sqrt 5+1}{4}}$

$\begin{aligned} \mathscr X & = \tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ \\ & = \tan 9^\circ + \tan 81^\circ - \tan 27^\circ - \tan 63^\circ \\ & = \tan (45^\circ-36^\circ) + \tan (45^\circ+36^\circ) - \tan (45^\circ-18^\circ) - \tan (45^\circ+18^\circ) \\ & = \frac {1 - \tan 36^\circ}{1 + \tan 36^\circ} + \frac {1 + \tan 36^\circ}{1 - \tan 36^\circ} - \frac {1 - \tan 18^\circ}{1 + \tan 18^\circ} - \frac {1 + \tan 18^\circ}{1 - \tan 18^\circ} \\ & = \frac {(1 - \tan 36^\circ)^2 + (1 + \tan 36^\circ)^2}{1 - \tan^2 36^\circ} - \frac {(1 - \tan 18^\circ)^2 + (1 + \tan 18^\circ)^2}{1 - \tan^2 18^\circ} \\ & = \frac {2(1 + \tan^2 36^\circ)}{2(1 - \tan^2 36^\circ)} - \frac {2(1 + \tan^2 18^\circ)}{2(1 - \tan^2 18^\circ)} \quad \quad \small \color{#3D99F6}{\text{Note that: }\cos 2 \theta = \frac {1-\tan^2 \theta}{1+ \tan^2 \theta}} \\ & = \frac 2{\cos 72^\circ} - \frac 2{\cos 36^\circ} \\ & = \frac {2(\cos 36^\circ \color{#3D99F6}{- \cos 72^\circ})}{\cos 72^\circ \cos 36^\circ} \quad \quad \small \color{#3D99F6}{\text{Note that: }\cos (180^\circ - \theta) = - \cos \theta} \\ & = \frac {2(\cos 36^\circ \color{#3D99F6}{+ \cos 108^\circ})}{\cos 72^\circ \cos 36^\circ} \quad \quad \small \color{#3D99F6}{\text{Note that: }\sum_{k=0}^{n-1} \cos \frac {(2k+1)\pi}{(2n+1)} = \frac 12} \\ & = \frac 1{\cos 72^\circ \cos 36^\circ} \\ & = \frac {\sin 36^\circ}{\cos 72^\circ \cos 36^\circ \sin 36^\circ} \\ & = \frac {2 \sin 36^\circ}{\cos 72^\circ \sin 72^\circ} \\ & = \frac {4 \color{#3D99F6}{\sin 36^\circ}}{\color{#3D99F6}{\sin 144^\circ}} \quad \quad \small \color{#3D99F6}{\text{Note that: }\sin (180^\circ - \theta) = - \sin \theta} \\ & = \frac {4 \color{#3D99F6}{\sin 36^\circ}}{\color{#3D99F6}{\sin 36^\circ}} \\ & = \boxed{4} \end{aligned}$