Trigonometry Problem 2

$\begin{aligned} a+b & = \sin 10^\circ + \sin 50^\circ \\ & = \sin (30^\circ -20^\circ) + \sin (30^\circ + 20^\circ) \\ & = \sin 30^\circ \cos 20^\circ - \sin 20^\circ \cos 30^\circ + \sin 30^\circ \cos 20^\circ + \sin 20^\circ \cos 30^\circ \\ & = 2\sin 30^\circ \cos 20^\circ \\ & = 2\cdot \frac 12 \cdot \sin (90^\circ - 20^\circ) \\ & = \sin 70^\circ \\ & = c \end{aligned}$

$\begin{aligned} abc & = \sin 10^\circ \sin 50^\circ \sin 70^\circ \\ & = \cos 80^\circ \cos 40^\circ \cos 20^\circ \\ & = \frac {\sin 20^\circ \cos 20^\circ \cos 40^\circ \cos 80^\circ}{\sin 20^\circ} \\ & = \frac {\sin 40^\circ \cos 40^\circ \cos 80^\circ}{2\sin 20^\circ} \\ & = \frac {\sin 80^\circ \cos 80^\circ}{4\sin 20^\circ} \\ & = \frac {\sin 160^\circ}{8\sin 20^\circ} & \small {\color{#3D99F6}\sin (180^\circ - \theta) = \sin \theta} \\ & = \frac {\sin 20^\circ}{8\sin 20^\circ} \\ & = \frac 18 \end{aligned}$

$\begin{aligned} a + b - c & = 0 \\ (a + b - c)^2 & = 0 \\ a^2+b^2+c^2 + 2(ab-bc-ca) & = 0 \\ bc + ca - ab & = \frac 12(a^2+b^2+c^2) \\ & = \frac 12 ( \sin^2 10^\circ + \sin^2 50^\circ + \sin^2 70^\circ) \\ & = \frac 14 (3 - \cos 20^\circ - \cos 100^\circ - \cos 140^\circ) \\ & = \frac 14 (3 - \sin 70^\circ + \sin 10^\circ + \sin 50^\circ) \\ & = \frac 14 (3 - 0) \\ & = \frac 34 \end{aligned}$

Now, we have:

$\begin{aligned} X & = 8abc \left(\frac {a+b}c \right) \left(\frac 1a+\frac 1b - \frac 1c \right) \\ & = 8 \cdot \frac 18 \left(\frac cc \right) \left(\frac {bc+ca-ab}{abc} \right) \\ & = 8 \cdot \frac 18 \cdot 1 \cdot \frac 34 \cdot 8 \\ & = \boxed{6} \end{aligned}$