ABCD is a trapezium such that AB and CD are parallel and BC ⊥ CD ....

Let $∠BDC = \alpha$, then

$tan\alpha = \dfrac{BC}{CD} = \dfrac{p}{q}$

$AE = BC = p$

$\tan \left(\alpha + \theta\right) = \dfrac{AE}{ED}$

$⇒ ED = \dfrac{p}{\tan \left(\alpha + \theta\right)}$

$= \dfrac{p\left(1 - tan\alpha tan\theta\right)}{tan\alpha + tan\theta}$

$= \dfrac{p\left(1 - \dfrac{p}{q}tan\theta\right)}{\dfrac{p}{q} + tan\theta}$

Now, $AB = q - ED$

$= q - \dfrac{p\left(q - ptan\theta\right)}{\dfrac{p}{q} + qtan\theta}$

$= \dfrac{pq + {q}^{2}tan\theta - pq + {p}^{2}tan\theta}{p + qtan\theta}$

$= \dfrac{\left({p}^{2} + {q}^{2}\right)tan\theta}{p + qtan\theta}$

$= \dfrac{\left({p}^{2} + {q}^{2}\right)sin\theta}{pcos\theta + qsin\theta}$