$ABCD$ is a trapezium such that $AB$ and $CD$ are parallel and $BC \perp CD$. If $\angle ADB = \theta, BC = p$ and $CD = q,$ then $AB$ is equal to :

Question

$\begin{array}{1 1} (1) \large\frac{(p^2+q^2)\sin \theta}{p \cos \theta+q \sin \theta } \\(2)\large\frac{p^2+q^2 \cos \theta}{p \cos \theta+q \sin \theta } \\(3)\large\frac{p^2+q^2}{p^2 \cos \theta + q^2 \sin \theta } \\ (4) \large\frac{(p^2+q^2)\sin \theta}{(p \cos \theta +q \sin \theta)^2} \end {array}$