If sin 21^{circ} = frac{x}{y}, then sec 21^{c | vedclass

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(A) Given $\sin 21^{\circ} = \frac{x}{y}.$Since $\cos 21^{\circ} = \sqrt{1 - \sin^{2} 21^{\circ}},$ we have $\cos 21^{\circ} = \sqrt{1 - \frac{x^{2}}{

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

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(A) Given $\sin 21^{\circ} = \frac{x}{y}.$Since $\cos 21^{\circ} = \sqrt{1 - \sin^{2} 21^{\circ}},$ we have $\cos 21^{\circ} = \sqrt{1 - \frac{x^{2}}{y^{2}}} = \frac{\sqrt{y^{2}-x^{2}}}{y}.$Now,$\sec 21^{\circ} = \frac{1}{\cos 21^{\circ}} = \frac{y}{\sqrt{y^{2}-x^{2}}}.$Also,$\sin 69^{\circ} = \sin(90^{\circ} - 21^{\circ}) = \cos 21^{\circ} = \frac{\sqrt{y^{2}-x^{2}}}{y}.$Therefore,$\sec 21^{\circ} - \sin 69^{\circ} = \frac{y}{\sqrt{y^{2}-x^{2}}} - \frac{\sqrt{y^{2}-x^{2}}}{y}.$Taking the common denominator $y\sqrt{y^{2}-x^{2}},$ we get:$= \frac{y^{2} - (y^{2}-x^{2})}{y\sqrt{y^{2}-x^{2}}} = \frac{x^{2}}{y\sqrt{y^{2}-x^{2}}}.$

If $\sin 21^{\circ} = \frac{x}{y},$ then $\sec 21^{\circ} - \sin 69^{\circ}$ is equal to

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