Discussion
If \(\frac{\cos A}{\cos B} = \alpha\), then \(\frac{\alpha + 1}{\alpha - 1}\) is equal to
Answer: D
\(\frac{\alpha + 1}{\alpha - 1}\) equals \(-\cot\left(\frac{A+B}{2}\right)\cot\left(\frac{A-B}{2}\right)\).
Substitute \(\alpha = \frac{\cos A}{\cos B}\) and multiply through by \(\cos B\), then use the sum-to-product formulas:
\[ \begin{aligned} \frac{\alpha + 1}{\alpha - 1} &= \frac{\cos A + \cos B}{\cos A - \cos B} \\ &= \frac{2\cos\frac{A+B}{2}\cos\frac{A-B}{2}}{-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}} \end{aligned} \]
Each cosine over the matching sine is a cotangent, and the minus sign from \(\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\) stays in front.
The first option loses that minus sign. The options with a tangent pair a cosine with the wrong sine, and the last drops the second factor entirely.
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