If ${\sin }^2\left(10°\right)\sin \left(20°\right)\sin \left(40°\right)\sin \left(50°\right)\sin \left(70°\right)=\alpha -$ $\frac{1}{16}\sin \left(10°\right)$, then $16+{\alpha }^{-1}$ is equal to _____.

We have, ${\sin }^{2}10°\times \sin 20°\times \sin 40°\times \sin 50°\times \sin 70°$

$=\left(\sin 10°\times \sin 50°\times \sin 70°\right)\left(\sin 10°\times \sin 20°\times \sin 40°\right)$

Using, $\sin \theta =\cos \left(90°-\theta \right)$

$= \left[\cos 80°\times \cos 40°\times \cos 20°\right]\left[\sin 10°\times \sin \left(30°-10°\right)\times \sin \left(30°+10°\right)\right]$

We know that $\cos \theta \cos 2\theta \cos 2^2\theta \dots \cos 2^{n-1}\theta =\frac{\sin 2^n\theta }{2^n\sin \theta }$

and $\sin \left(A+B\right)·\sin \left(A-B\right)={\sin }^{2}A-{\sin }^{2}B$$\Rightarrow \left(\frac{\sin 2^3\left(20°\right)}{2^3\sin 20°}\right)\left[\sin 10°\times \left(\frac{1}{4}-{\sin }^{2}10°\right)\right]$

$\Rightarrow \left(\frac{\sin 160°}{2^3\sin 20°}\right)\left(\frac{\sin 10°-4{\sin }^{3}10°}{4}\right)$

$\Rightarrow \left(\frac{\sin \left(180°-20°\right)}{8\sin 20°}\right)\left(\frac{3\sin 10°-4{\sin }^{3}10°-2\sin 10°}{4}\right)$

$\Rightarrow \left(\frac{\sin 20°}{8\sin 20°}\right)\left(\frac{\sin 30°-2\sin 10°}{4}\right)$

$\Rightarrow \frac{1}{8}\left(\frac{\frac{1}{2}-2\sin 10°}{4}\right)$

$\Rightarrow \frac{1}{8}\left(\frac{1-4\sin 10°}{8}\right)=\alpha -\frac{\sin 10°}{16}$   (given)

$\Rightarrow \frac{1-4\sin 10°}{64}=\frac{16\alpha -\sin 10°}{16}$

$\Rightarrow 1-4\sin 10°=64\alpha -4\sin 10°$

$\Rightarrow \alpha =\frac{1}{64}$.