已知(1+tanA)/(1-tanA)=3+2根号2,求证:[(sinA+cosA)^2-1]/(cotA-sinAco

(1+tanA)/(1-tanA)=3+2根号2
tanA=1/2*根号2
[(sinA+cosA)^2-1]/(cotA-sinAcosA)
=(sin^2A+cos^2A+2sinAcosA-1)/[(cosA-sin^2AcosA)/sinA]
=2sinAcosA/[cosA(1-sin^2A)/sinA]
=2sinAcosA/[cosAcos^2A)/sinA]
=2tan^2A
=1