A trigonometric problem

Relevant wiki: Sum and Difference Trigonometric Formulas - Problem Solving

Consider

$\begin{aligned} x^2 - 12x + 32 & = 0 \\ (x-4)(x-8) & = 0 \\ \implies \tan A, \tan B & = 4, 8 \\ \implies \tan (A+B) & = \frac {4+8}{1-(4)(8)} = - \frac {12}{31} \end{aligned}$

Now, we have:

$\begin{aligned} X & = \sin^2 (A+B) - 12 \sin(A+B) \cos(A+B) + 32 \cos^2 (A+B) \\ & = \cos^2 (A+B) \left[\tan^2 (A+B) - 12 \tan(A+B) + 32 \right] \\ & = \frac {(\tan (A+B) -4)(\tan (A+B) - 8)}{\sec^2 (A+B)} \\ & = \frac {\left(- \frac {12}{31}-4\right)\left(- \frac {12}{31}-8\right)}{1+\left(- \frac {12}{31}\right)^2} \\ & = \frac {(136)(260)}{1105} \\ & = \boxed{32} \end{aligned}$