Sin 53

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28 cos A + 45 sin A = 53

53 sin A = 55

Prove that : $(sin 55^{\circ} - sin 19^{\circ}) + (sin 53^{\circ} - sin 17^{\circ}) = cos 1^{\circ}$

({cos}^{2} 37^{\circ} - {sin}^{2} 53^{\circ}) =

\frac{{sin}^{2} 53 + {cos}^{2} 53}{{sec}^{2} 37 - {tan}^{2} 37}

sin 55^{o} - sin 19^{o} + sin 53^{o} - sin 17^{o}

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