Tangent Tango

Because $f(1) = A + B + C + D + 1 = (1 - \tan^2(12^\circ))(1 - \tan^2(24^\circ))(1 - \tan^2(48^\circ))(1 - \tan^2(96^\circ))$ .

Also, by double angle identity, $1- \tan^2 (X) = \frac{\cos^2 (X) - \sin^2(X)}{\cos^2(X)} = \frac{\cos(2X)}{\cos^2(X)}$ . Then,

$\begin{array} {r c l} A + B + C + D + 1 & =& \dfrac{\cos(24^\circ)}{\cos^2(12^\circ)} \cdot \dfrac{\cos(48^\circ)}{\cos^2(24^\circ)} \cdot \dfrac{\cos(96^\circ)}{\cos^2(48^\circ)} \cdot \dfrac{\cos(192^\circ)}{\cos^2(96^\circ)} \\ \phantom0\\ &=& \dfrac{-\cos(12^\circ)}{\cos^2(12^\circ) \cos(24^\circ) \cos(48^\circ) \cos(96^\circ)} \\ \phantom0\\ &=& -\dfrac{1}{\cos(12^\circ) \cos(24^\circ) \cos(48^\circ) \cos(96^\circ)} \end{array}$

Also, with $\sin(2X) = 2\sin(X) \cos(X)$ .

Multiplying the expression above by $\dfrac{16\sin(12^\circ) \sin(24^\circ) \cos(48^\circ) \cos(96^\circ)}{16\sin(12^\circ) \sin(24^\circ) \cos(48^\circ) \cos(96^\circ)}$ simplifies to $A+B+C+D+1=16$ .

The answer is $16 - 1 =\boxed{15} .$

Note that $f(1) = 1 + A + B + C + D$ and that $f(x) = (x - \tan^2 12^{\circ} )(x - \tan^2 24^{\circ} )(x - \tan^2 48^{\circ} )(x - \tan^2 96^{\circ} )$ . So plugging in $x = 1$ gives $f(1) = 16$ . Subtracting $1$ from that gives the answer of $\boxed{15}$