Concept:We use the identity sin2θ+cos2θ=1 to express sin2α and cos2α in terms of p, q, and m. Similarly for β using n. Then tan2θ=cos2θsin2θ.Explanation:Start with psin2α+qcos2α=m.Replace cos2α=1−sin2α: psin2α+q(1−sin2α)=m.Simplify: (p−q)sin2α=m−q.So sin2α=p−qm−q.Then cos2α=1−sin2α=p−qp−m.Thus tan2α=cos2αsin2α=p−mm−q.Now for β: qsin2β+pcos2β=n.Substitute cos2β=1−sin2β: qsin2β+p(1−sin2β)=n.Simplify: (q−p)sin2β=n−p.Hence sin2β=q−pn−p.Then cos2β=q−pq−n.Therefore tan2β=q−nn−p.Now (tanβtanα)2=tan2βtan2α=p−mm−q×n−pq−n.Notice p−m=−(m−p) and q−n=−(n−q). Also n−p=−(p−n). Simplify signs: the two negatives from p−m and q−n cancel, leaving (m−p)(n−q)(m−q)(n−p).Answer:(m−p)(n−q)(m−q)(n−p), which matches option D.