# trigonometry – If \$ csc theta = sqrt { frac {p + q} {pq}} \$ where \$ p> q> 0 \$, then \$ | cot ( frac { pi} {4} + frac { theta} {2}) | \$ is equal to

Yes $$csc theta = sqrt { frac {p + q} {p-q}}$$ or $$p> q> 0$$then $$| cot ( frac { pi} {4} + frac { theta} {2}) |$$ is equal to:

What I've tried

Since, $$cot ( frac { pi} {4} + frac { theta} {2}) = sqrt { frac {1+ cos ( frac { pi} {2} + theta) } {1- cos ( frac { pi} {2} + theta)}}$$
$$= sqrt { frac {1- sin theta} {1 + sin theta}}$$
$$= sqrt { frac {1- frac {p-q} {p + q}} {1+ frac {p-q} {p + q}}}}$$
$$= sqrt { frac {q} {p}}$$

That's what I have but the answer is $$sqrt { frac {p} {q}}$$. I do not know where I did wrong. Please help! Thank you!