Question

化简：sin^2`x/(sinx-cosx)- (sinx+cosx)/tan^2`x - 1.

Answer 1

通分,分母＝sin^2`x（tan^2`x - 1）-(sinx+cosx)(sinx-cosx)＝(sin^2-cos^2)^2/cos^2　中间步骤略

分子＝(sin^2-cos^2)(sin-cos)/cos^2

化简得sinx+cosx