What is \frac{\tan^2 \alpha}{\tan^2 \beta} equal to?
For the following two (02) items: Let p\sin^2 \alpha + q\cos^2 \alpha = m, q\sin^2 \beta + p\cos^2 \beta = n; p \neq m, n and q \neq m, n.
- A. -\frac{(m-q)(n-q)}{(m-p)(n-p)}
- B. -\frac{(m-q)(n-p)}{(m-p)(n-q)}
- C. \frac{(m-q)(n-q)}{(m-p)(n-p)} ✓
- D. \frac{(m-q)(n-p)}{(m-p)(n-q)}
Correct Answer: C. \frac{(m-q)(n-q)}{(m-p)(n-p)}
Explanation
From p\sin^2 \alpha + q\cos^2 \alpha = m, dividing by \cos^2 \alpha yields p\tan^2 \alpha + q = m\sec^2 \alpha = m(1+\tan^2 \alpha), which gives \tan^2 \alpha = \frac{m-q}{p-m}. Similarly, \tan^2 \beta = \frac{n-p}{q-n}. Their ratio is \frac{(m-q)(n-q)}{(m-p)(n-p)}.
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