C : 2816

The problem was published in GMB 12 / 2004; I don't think it's here. It was solved in GMB 6 / 2005.

    "C:2816.  Prove the equality  $\frac{\sin 110^{\circ}}{\sin 30^{\circ}}-\frac{\sin 60^{\circ}}{\sin 80^{\circ}}=1."$

{authors :  Romanța GHIȚĂ and Ioan GHIȚĂ, Blaj

               Solution CiP - taken from the cited source

          The given equality is successively equivalent to :

$\sin 110^{\circ}\cdot \sin 80^{\circ}=\sin 30^{\circ}\cdot \sin 60^{\circ}+\sin 30^{\circ}\cdot \sin 80^{\circ}\;\Leftrightarrow$

$\Leftrightarrow\;\;2\sin 70^{\circ}\cdot \sin 80^{\circ}=\sin 60^{\circ}+\sin 80^{\circ}\;\Leftrightarrow$

$\Leftrightarrow\;\;\cos 10^{\circ}-\cos 150^{\circ}=\sin 60^{\circ}+\sin 80^{\circ}$

which is obvious, because  $\cos 10^{\circ}=\sin 80^{\circ}\;\;and\;\; -\cos 150^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}.$

$\blacksquare$

          REMARK CiP   It would be interesting to study some trigonometric equations starting from the present equality, as I did in the post of September/7/ 2025  A forgotten trigonometric equation.