Question

已知tana=-1/2,则sin^a+sinacosa-cos^a的值为？

请带有详细步骤。。。谢谢

Answer 1

巧解
sin^a+sinacosa-cos^a
=（sin^a+sinacosa-cos^a）/(sin^a++cos^a) 上下同除以cos^a
=(tan^a+tana-1)/(tan^a+1)
=(1/4-1/2-1)/(1/4+1)
=-1

Answer 2

sec^2a=1+tg^2a=5/4
cos^2a=4/5
原式=[（sin^a+sinacosa-cos^a)/cos^2a ]*cos^2a
=(tg^2a+tga-1)*cos^2a =(1/4-1/2-1)*4/5=-1