psin²α + qcos²α = m and pcos²β + qsin²β = n. What is tan²α equal to?

From psin²α + qcos²α = m, write psin²α + q(1−sin²α) = m, so (p−q)sin²α = m−q, giving sin²α = (m−q)/(p−q). Similarly cos²α = (p−m)/(p−q). Therefore tan²α = sin²α/cos²α = (m−q)/(p−m). This matches option (c) written as (m−q)/(m−p)? Reexamining: (m−q)/(p−m) — option (a) shows (m−q)/(p−m). So the correct answer is option (a).