Q.If \(\sin 17^\circ = \frac{x}{y}\), then show that \(\sec 17^\circ - \sin 73^\circ = \frac{x^2}{y\sqrt{y^2 - x^2}}\)

Given: \(\sin 17^\circ = \frac{x}{y}\) ⇒ \(\sin^2 17^\circ = \left(\frac{x}{y}\right)^2\) ⇒ \(1 - \sin^2 17^\circ = 1 - \frac{x^2}{y^2}\) ⇒ \(\cos^2 17^\circ = \frac{y^2 - x^2}{y^2}\) ⇒ \(\cos 17^\circ = \sqrt{\frac{y^2 - x^2}{y^2}} = \frac{\sqrt{y^2 - x^2}}{y}\) Now consider: \(\sec 17^\circ - \sin 73^\circ\) \(= \frac{1}{\cos 17^\circ} - \frac{1}{\sin(90^\circ - 17^\circ)}\) \(= \frac{1}{\cos 17^\circ} - \cos 17^\circ\) \(= \frac{1}{\frac{\sqrt{y^2 - x^2}}{y}} - \frac{\sqrt{y^2 - x^2}}{y}\) \(= \frac{y}{\sqrt{y^2 - x^2}} - \frac{\sqrt{y^2 - x^2}}{y}\) Now simplify: \(= \frac{y^2 - (y^2 - x^2)}{y\sqrt{y^2 - x^2}}\) \(= \frac{y^2 - y^2 + x^2}{y\sqrt{y^2 - x^2}}\) \(= \frac{x^2}{y\sqrt{y^2 - x^2}}\) ( proved).
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