Discussion
If \(\sin 17^{\circ}=\frac{a}{b}\), then the value of \(\sec 17^{\circ} - \sin 73^{\circ}\) is :
Answer: A
The value is \(\frac{a^{2}}{b\sqrt{b^{2}-a^{2}}}\).
Since \(\sin 73^{\circ} = \cos 17^{\circ}\), the expression is \(\sec 17^{\circ} - \cos 17^{\circ}\). With \(\sin 17^{\circ} = a/b\), \(\cos 17^{\circ} = \sqrt{b^{2}-a^{2}}/b\):
\[ \begin{aligned} \sec\theta - \cos\theta &= \frac{1 - \cos^{2}\theta}{\cos\theta} = \frac{\sin^{2}\theta}{\cos\theta} \\ &= \frac{a^{2}/b^{2}}{\sqrt{b^{2}-a^{2}}/b} \\ &= \frac{a^{2}}{b\sqrt{b^{2}-a^{2}}} \end{aligned} \]
The options with \(\sqrt{a^{2}-b^{2}}\) cannot be right, because \(a < b\) for a sine and that root would not be real. The option with \(b^{2}\) on top inverts the sine.
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