证(1/sin^2x)+(1/cos^2x)-(1/tan^2x)=2+tan^2x附加题（10分）：(1-sin^4x-cos^4x)/(sin^2x-sin^4x)
问题描述：
证(1/sin^2x)+(1/cos^2x)-(1/tan^2x)=2+tan^2x
附加题（10分）：(1-sin^4x-cos^4x)/(sin^2x-sin^4x)
答
(1/sin^2x)+(1/cos^2x)-(1/tan^2x）
=（1 /sin^2xcos^2x)-(cos^2x/sin^2x)
=(1-cos^4x)/sin^2xcos^2x
=(1+cos^2x)/cos^2x
=1+1/cos^2x
=1+1+sin^2x/cos^2x
=2+tan^2x
答
(1/sin^2x)+(1/cos^2x)-(1/tan^2x)
=[(cosx)^2+(sinx)^2]/(sinxcosx)^2-(cosx)^2/(sinx)^2
=1/(sinxcosx)^2-(cosx)^4/(sinxcosx)^2
=[1-(cosx)^2][1+(cosx)^2]/(sinxcosx)^2
=(sinx)^2[1+(cosx)^2]/(sinxcosx)^2
=[1+(cosx)^2]/(cosx)^2
=[2(cosx)^2+(sinx)^2]/(cosx)^2
=2+(tanx)^2
(1-sin^4x-cos^4x)/(sin^2x-sin^4x)
={[1-(sinx)^2][1+(sinx)^2]-(cosx)^4}/{(sinx)^2[1-(sinx)^2]}
={(cosx)^2[1+(sinx)^2]-(cosx)^4}/(sinxcosx)^2
=(cosx)^2[1+(sinx)^2-(cosx)^2]/(sinxcosx)^2
=2(sinx)^2/(sinx)^2
=2