What is \cot^2\beta 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

  1. A. \frac{n-q}{p-n}
  2. B. \frac{n-p}{q-n}
  3. C. \frac{n-p}{n-q}
  4. D. \frac{n-q}{n-p}

Correct Answer: A. \frac{n-q}{p-n}

Explanation

Substitute \cos^2\beta = 1 - \sin^2\beta. The equation is p(1 - \sin^2\beta) + q\sin^2\beta = n \implies \sin^2\beta(q-p) = n-p \implies \sin^2\beta = \frac{p-n}{p-q}. Thus \cos^2\beta = 1 - \frac{p-n}{p-q} = \frac{n-q}{p-q}. Therefore, \cot^2\beta = \frac{\cos^2\beta}{\sin^2\beta} = \frac{n-q}{p-n}.

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