1. f(a)=[2sin( π+a)cos(π-a)-cos(π+a)]/[1+sin^2a+sin(π-a)-cos^2(π-a)]
=(2sinacosa+cosa)/(1+sin²a+sina-cos²a)
=cosa(2sina+1)/[sina(2sina+1)]
=cota
(1) a=-17π/6
f(a)=cot(-17π/6)=-cot(17π/6)=-cot(3π-π/6)
=cot(π/6)=√3
(2) 若a是銳角,且sin(a-3π/2)=3/5
即sin(3π/2-a)=-3/5
-cosa=-3/5 cosa=3/5
sina=√(1-cos²a)=4/5
所以f(a)=cota=cosa/sina=(3/5)/(4/5)=3/4
2. 已知sin&=3/5,&屬于（π/2,π）,tan&<0cos&<0
則cos&=-√(1-cos²&)=-4/5
tan&=sin&/cos&=(3/5)/(-4/5)=-3/4
希望能幫到你O(∩_∩)O