If ^ 2 (10 °) (20 °) (40 °) (50 °) (70 °) = α - (1)/(16) (10 °) ,...

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 $cos\theta cos2\theta cos{2}^{2}\theta \ldots cos{2}^{n - 1}\theta = \dfrac{sin{2}^{n}\theta}{{2}^{n}sin\theta}$

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

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

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

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

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

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

$⇒ \dfrac{1 - 4\sin 10 °}{64} = \dfrac{16\alpha - \sin 10 °}{16}$

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

$⇒ \alpha = \dfrac{1}{64}$.