If sin 17^{circ} = frac{x}{y}, then the value | vedclass

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(B) Given that $\sin 17^{\circ} = \frac{x}{y}.$Using the identity $\cos^{2} 	heta + \sin^{2} 	heta = 1,$ we find $\cos 17^{\circ} = \sqrt{1 - \sin^{

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$\frac{x^{2}}{y \sqrt{y^{2}-x^{2}}}$

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(B) Given that $\sin 17^{\circ} = \frac{x}{y}.$Using the identity $\cos^{2} 	heta + \sin^{2} 	heta = 1,$ we find $\cos 17^{\circ} = \sqrt{1 - \sin^{2} 17^{\circ}}.$$\cos 17^{\circ} = \sqrt{1 - \frac{x^{2}}{y^{2}}} = \sqrt{\frac{y^{2} - x^{2}}{y^{2}}} = \frac{\sqrt{y^{2} - x^{2}}}{y}.$Therefore,$\sec 17^{\circ} = \frac{1}{\cos 17^{\circ}} = \frac{y}{\sqrt{y^{2} - x^{2}}}.$Using the complementary angle identity $\sin(90^{\circ} - 	heta) = \cos 	heta,$ we have $\sin 73^{\circ} = \sin(90^{\circ} - 17^{\circ}) = \cos 17^{\circ} = \frac{\sqrt{y^{2} - x^{2}}}{y}.$Now,calculate $(\sec 17^{\circ} - \sin 73^{\circ}) = \frac{y}{\sqrt{y^{2} - x^{2}}} - \frac{\sqrt{y^{2} - x^{2}}}{y}.$Taking the common denominator: $\frac{y^{2} - (\sqrt{y^{2} - x^{2}})^{2}}{y \sqrt{y^{2} - x^{2}}} = \frac{y^{2} - (y^{2} - x^{2})}{y \sqrt{y^{2} - x^{2}}} = \frac{x^{2}}{y \sqrt{y^{2} - x^{2}}}.$

If $\sin 17^{\circ} = \frac{x}{y},$ then the value of $(\sec 17^{\circ} - \sin 73^{\circ})$ is

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