Golden Ratio related Trig simplification Show that
$$ \tan72^\circ - \cot36^\circ   - \cot54^\circ =  3 \tan18^\circ $$
Reference is to
TangentSubtr&Addn
EDIT1:
( It is not so easy or straightforward as it appears at first sight!)
 A: A little generalization:
Replacing $18^\circ$ with $t,$
$$\dfrac1{\tan t}-3\tan t=\dfrac{4\cos^2t-3}{\sin t\cos t}=\dfrac{2\cos3t}{\cos 
t\sin2t}=\dfrac{2\cos3t(2\cos2t)}{\cos t\sin4t}=\dfrac{2\{\cos(3t-2t)+\cos(3t+2t)\}}{\cos t\sin4t}$$
$$\implies\dfrac1{\tan t}-3\tan t=\dfrac2{\sin4t}$$
if  $\cos t\ne0$ and $\cos5t=0\iff5t=(2n+1)90^\circ$
$\iff t=(2n+1)18^\circ$ where $n\not\equiv2\pmod5$
Now $\tan2t+\dfrac1{\tan2t}=\dfrac2{\sin4t}$
Here $n=0$
A: Notice that all of the angles are multiples of 18. So it may help using the double, triple or quadruple angle formula:
$\tan2\theta \equiv \frac{2\tan\theta}{1-\tan^2\theta}$
$\tan3\theta \equiv \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$
$\tan4\theta \equiv \frac{4\tan\theta-4\tan^3\theta}{1-6\tan^2\theta+\tan^4\theta}$
http://mathworld.wolfram.com/Multiple-AngleFormulas.html
Also, noticing that $\tan(90^\circ-\theta) = \cot(\theta)$ may also help.
A: $$\cot36^\circ+\cot54^\circ=\dfrac2{\sin72^\circ}\  \ \ \  (1)$$
$$\tan72^\circ-\tan18^\circ=\dfrac{\sin(72-18)^\circ}{\cos18^\circ\sin18^\circ}=\dfrac{2\sin54^\circ}{\sin36^\circ}=2\cot36^\circ\  \ \ \  (2)$$
$$\cot36^\circ-\tan18^\circ=\cot36^\circ-\cot72^\circ=\dfrac{\sin(72-36)^\circ}{\sin36^\circ\sin72^\circ}=\dfrac1{\sin72^\circ}\  \ \ \  (3)$$
$$\implies\tan72^\circ - \cot36^\circ   - \cot54^\circ -3 \tan18^\circ$$
$$=(\underbrace{\tan72^\circ-\tan18^\circ})-(\underbrace{\cot36^\circ+\cot54^\circ})-2\tan18^\circ$$
$$=2(\underbrace{\cot36^\circ-\tan18^\circ})-\dfrac2{\sin72^\circ}(\text{using }(2),(1))$$
Now use $(3)$
