If $$\sin x + \sin^2 x = 1$$, $$x \in (0, \tfrac{\pi}{2})$$ then $$(\cos^{12}x+\tan^{12}x)+3(\cos^{10}x+\tan^{10}x+\cos^{8}x+\tan^{8} x)+(\cos^{6}x+\tan^{6}x)$$
is equal to :

Given: $$\sin x + \sin^2 x = 1$$, which means $$\sin x = 1 - \sin^2 x = \cos^2 x$$.

Also: $$\tan x = \frac{\sin x}{\cos x}$$, and since $$\sin x = \cos^2 x$$:

$$\tan^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{\sin^2 x}{\sin x} = \sin x = \cos^2 x$$

So $$\cos^2 x = \tan^2 x$$. Let $$t = \cos^2 x = \tan^2 x$$.

The expression becomes:

$$ (t^6 + t^6) + 3(t^5 + t^5 + t^4 + t^4) + (t^3 + t^3) $$

$$ = 2t^6 + 6t^5 + 6t^4 + 2t^3 = 2t^3(t^3 + 3t^2 + 3t + 1) = 2t^3(t+1)^3 $$

Now we need to find $$t$$. From $$\sin x = \cos^2 x$$ and $$\sin^2 x + \cos^2 x = 1$$:

$$\sin^2 x + \sin x = 1$$ (which is the given condition).

$$t = \cos^2 x = \sin x$$, and $$t + t^2 = 1$$, so $$t^2 + t = 1$$, i.e., $$t + 1 = \frac{1}{t}$$... actually:

$$t^2 + t - 1 = 0 \Rightarrow t = \frac{-1+\sqrt{5}}{2}$$ (taking positive root since $$x \in (0, \pi/2)$$).

So $$t(t+1) = t \cdot \frac{1}{t} \cdot t$$... Let me recalculate:

From $$t^2 + t = 1$$: $$t(t+1) = 1$$.

Therefore: $$2t^3(t+1)^3 = 2[t(t+1)]^3 = 2(1)^3 = 2$$.

The correct answer is Option 4: 2.