What is \tan^2\alpha equal to ?
For the next two (02) items that follow : p\sin^2\alpha + q\cos^2\alpha = m and p\cos^2\beta + q\sin^2\beta = n
- A. \frac{m-q}{p-m} ✓
- B. \frac{m-p}{q-m}
- C. \frac{m-q}{m-p}
- D. \frac{m-p}{m-q}
Correct Answer: A. \frac{m-q}{p-m}
Explanation
Rewrite q\cos^2\alpha as q(1 - \sin^2\alpha). The equation becomes p\sin^2\alpha + q - q\sin^2\alpha = m \implies \sin^2\alpha (p-q) = m-q \implies \sin^2\alpha = \frac{m-q}{p-q}. Then \cos^2\alpha = 1 - \sin^2\alpha = 1 - \frac{m-q}{p-q} = \frac{p-m}{p-q}. Therefore, \tan^2\alpha = \frac{\sin^2\alpha}{\cos^2\alpha} = \frac{m-q}{p-m}.
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