psin²α + qcos²α = m and pcos²β + qsin²β = n. What is cot²β equal to?

From pcos²β + qsin²β = n, p(1−sin²β) + qsin²β = n, so (q−p)sin²β = n−p, giving sin²β = (n−p)/(q−p). Also cos²β = (q−n)/(q−p). Therefore cot²β = cos²β/sin²β = (q−n)/(n−p) = (n−p)/(q−n) when sign-adjusted? Specifically (q−n)/(n−p) equals option (b) (n−p)/(q−n)? No — they are reciprocals. Recompute: cot²β = cos²β/sin²β = [(q−n)/(q−p)] / [(n−p)/(q−p)] = (q−n)/(n−p). Multiplying numerator and denominator by −1: (n−q)/(p−n). This matches option (a) (n−q)/(p−n).