Trigonometry Identity

We first prove the trigonometric identity: $\cot x - \tan x = 2 \cot (2x)$

$\begin{aligned} \cot x - \tan x & = & \frac { \cos x}{\sin x } - \frac { \sin x}{\cos x } \\ & = & \frac {\cos^2 x - \sin^2 x }{\sin x \cos x } \\ & = & 2 \frac {\cos^2 x - \sin^2 x }{2 \sin x \cos x } \\ & = & 2 \frac {\cos (2x) }{\sin (2x)} \\ & = & 2 \cot (2x) \\ \end{aligned}$

By Binomial expansion

$\begin{aligned} (a - b)^4 & = & a^4 - 4a^3 b + 6a^2 b^2 - 4ab^3 + b^4 \\ & = & a^4+b^4 - 4ab(a^2+b^2) + 6a^2 b^2 \\ & = & a^4+b^4 - 4ab \left ( (a-b)^2 + 2ab \right ) + 6a^2 b^2 \\ & = & a^4 + b^4 - 4ab(a-b)^2 - 2a^2 b^2 \\ \end{aligned}$

$\begin{aligned} a^4+b^4 & = &(a-b)^4 + 4ab(a-b)^2 + 2a^2 b^2 \end{aligned}$

Let $a = \cot (11.25^\circ), b = \tan (11.25^\circ) \Rightarrow a - b = 2\cot (22.5^\circ), ab = 1$

$\Rightarrow \cot^4 (11.25^\circ) + \tan^4 (11.25^\circ) = 16 \cot^4 (22.5^\circ) + 16 \cot^2 (22.5) + 2$

Similarly, let $a = \cot (22.5^\circ), b = \tan (22.5^\circ)$

$\begin{aligned} \Rightarrow \cot^4 (22.5^\circ) + \tan^4 (22.5^\circ) & = & 16 \cot^4 (45^\circ) + 16 \cot^4 (45^\circ) + 2 \\ & = & 16 + 16+2=34 \end{aligned}$

Again, let $a = \cot (33.75^\circ), b = \tan (33.75^\circ)$

$\Rightarrow \cot^4 (33.75^\circ) + \tan^4 (33.75^\circ) = 16 \cot^4 (67.5^\circ) + 16 \cot^2 (67.5) + 2$

Hence,

$\cot^4 (11.25^\circ) +\tan^4 (11.25^\circ) + \cot^4 (22.5^\circ)$

$+ \tan^4 (22.5^\circ) + \cot^4 (33.75^\circ) + \tan^4 (33.75^\circ)$

$= \left ( 16 \cot^4 (22.5^\circ) + 16 \cot^2 (22.5) + 2 \right ) + 34 + \left ( 16 \cot^4 (67.5^\circ) + 16 \cot^2 (67.5) + 2 \right )$

$= 16 \left ( \cot^4 (22.5^\circ) + \cot^4 (67.5^\circ) \right ) + 16 \left ( \cot^2 (22.5^\circ) + \cot^2 (67.5^\circ) \right ) + 38$

$= 16 \left ( \cot^4 (22.5^\circ) + \cot^4 (22.5^\circ) \right ) + 16 \left ( \cot^2 (22.5^\circ) + \cot^2 (22.5^\circ) \right ) + 38$

$= 16(34) + 16 \left ( (\cot (22.5^\circ) - \tan (22.5^\circ)^2 + 2 \right ) + 582$

$= 16 \left ( (2 \cot (45^\circ) )^2 + 2 \right ) + 582$

$= 16(2^2 + 2) + 582$

$= \boxed{678}$