已知sinα=asinβ,bcosα=acosβ,且α,β为锐角,求证：cosα=√(a^2-1)/(b^2-1)

证明：sinα=asinβ,bcosα=acosβ,(sinα)^2=a^2(sinβ)^2,b^2(cosα)^2=a^2(cosβ)^2两式相加,1-(cosα)^2+b^2(coaα)^2=a^2(cosα)^2=(a^2-1)/(b^2-1) (b^2-1≠0)α是锐角,cosα=√(a^2-1)/(b^2-1)