If \tan \theta=\frac{\sqrt{q^{2}-p^{2}}}{p}; 0 \lt \theta \lt 90^{\circ}, then what is \sec \theta+\cos \theta+2 equal to?

  1. A. \frac{p^{2}+q^{2}}{pq}
  2. B. \frac{(p+q)^{2}}{pq} ✓
  3. C. \frac{(p+q)^{2}}{2pq}
  4. D. \frac{(p-q)^{2}}{pq}

Correct Answer: B. \frac{(p+q)^{2}}{pq}

Explanation

Given \tan\theta = \frac{\sqrt{q^2-p^2}}{p}, the hypotenuse is \sqrt{(\sqrt{q^2-p^2})^2+p^2} = q. Thus, \sec\theta = \frac{q}{p} and \cos\theta = \frac{p}{q}. The expression \sec\theta + \cos\theta + 2 = \frac{q}{p} + \frac{p}{q} + 2 = \frac{p^2+q^2+2pq}{pq} = \frac{(p+q)^2}{pq}.

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