If \csc \theta-\cot \theta=m and \sec \theta-\tan \theta=n, then what is \csc \theta+\sec \theta equal to?

  1. A. \frac{1}{2}(m+n+\frac{1}{m}+\frac{1}{n})
  2. B. (m+n+\frac{1}{m}+\frac{1}{n})
  3. C. \frac{1}{2}(m+n-\frac{1}{m}-\frac{1}{n})
  4. D. (m+n-\frac{1}{m}-\frac{1}{n})

Correct Answer: A. \frac{1}{2}(m+n+\frac{1}{m}+\frac{1}{n})

Explanation

We know \csc \theta + \cot \theta = \frac{1}{m} and \sec \theta + \tan \theta = \frac{1}{n}. Adding the conjugate pairs: (\csc \theta - \cot \theta) + (\csc \theta + \cot \theta) = m + \frac{1}{m} \implies 2\csc \theta = m + \frac{1}{m}. Similarly, 2\sec \theta = n + \frac{1}{n}. Adding these yields 2(\csc \theta + \sec \theta) = m + n + \frac{1}{m} + \frac{1}{n}.

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