If $\sin x + \sin^2 x = 1$, $x \in \left(0, \frac{\pi}{2}\right)$, 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:

Let the range of the function $f(x)=6+16 \cos x \cdot \cos \left(\frac{\pi}{3}-x\right) \cdot \cos \left(\frac{\pi}{3}+x\right) \cdot \sin 3 x \cdot \cos 6 x, x \in \mathbf{R}$ be $[\alpha, \beta]$. Then the distance of the point $(\alpha, \beta)$ from the line $3 x+4 y+12=0$ is :

The value of $\left(\sin 70^{\circ}\right)\left(\cot 10^{\circ} \cot 70^{\circ}-1\right)$ is