If  $si{n}^2\left(10^0\right)sin\left(20^0\right)sin\left(40^0\right)sin\left(50^0\right)sin\left(70^0\right)=\alpha -\frac{1}{16}sin\left(10^0\right),$  then 16 + a^-1 is equal to………..

$\left(sin10°.sin50°.sin70°\right).\left(sin10°.sin20°.sin40°\right)$

$=\left(\frac{1}{4}sin30°\right).\left[\frac{1}{2}sin10°\left(cos20°-cos60°\right)\right]$

$=\frac{1}{32}\left[sin30°-sin10°-sin10°\right]$

$\frac{1}{64}-\frac{1}{16}sin10°$

Clearly $\alpha =\frac{1}{64}$  

              Hence 16 + a^-1 = 80