Trigonometry Manipulation

$\frac{sin x}{cos x}$ + $\frac{cos x}{sin x}$ = $\frac{sin^2x + cos^2x}{sinxcosx}$ = $\frac{1}{sinxcosx} = 12$ ; $sinxcosx$ = $\frac{1}{12}$ ; (Note : $sin 2x = 2sinxcosx$ ) -> $2sinxcosx = 1/6$ -> $sin2x = 1/6$ . Thus the answer is 1/6

Relevant wiki: Half Angle Tangent Substitution

By Weierstrass substitution or tangent half-angle substitution, we have:

$\begin{aligned} \sin 2x & = \frac {2\tan x}{1+\tan^2 x} & \small \color{#3D99F6} \text{Divide up and down of the RHS by } \tan x \\ & = \frac 2 {\cot x+\tan x} \\ & = \frac 2{12} = \boxed{\dfrac 16} \end{aligned}$