作业帮(1)已知COSa=-4/5,a属于(派,3派/2),求tan a.(2)若tan a=2,求(sin a+cos a)/(sin a-cos a) + cos^2 a的值.(注:^2为平方)
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/08 03:16:53
(1)已知COSa=-4/5,a属于(派,3派/2),求tan a.(2)若tan a=2,求(sin a+cos a)/(sin a-cos a) + cos^2 a的值.(注:^2为平方)
(1)已知COSa=-4/5,a属于(派,3派/2),求tan a.
(2)若tan a=2,求(sin a+cos a)/(sin a-cos a) + cos^2 a的值.(注:^2为平方)
(1)已知COSa=-4/5,a属于(派,3派/2),求tan a.(2)若tan a=2,求(sin a+cos a)/(sin a-cos a) + cos^2 a的值.(注:^2为平方)
(1)已知cosα=-4/5,a∈(π,3π/2),求tan a.
因为cosα=-4/5,所以secα=-5/4,a∈(π,3π/2),故tanα=√(sec²α-1)=√(25/16-1)=3/4；
2)若tanα=2,求(sin α+cos α)/(sin α-cos α) + cos²α的值
(sin α+cos α)/(sin α-cos α) + cos²α=（tanα+1)/(tanα-1)+1/(1+tan²α)
=(2+1)/(2-1)+1/(1+2²)=3+1/5=16/5.
a属于(派，3派/2)
sina<0
sina=-√(1-cos^2a)=-3/5
tana=sina/cosa=3/4
(2)
tana=2
sina=2cosa 平方一下
sin^2a=1-cos^2a=4cos^2a
cos^2a=1/5
(sina+cosa)/(sina-cosa)+cos^2 a
cosa(tana+1)/cosa(tana-1)+cos^2 a
=(tana+1)/(tana-1)+cos^2 a
=(2+1)/(2-1)+1/5
=3+1/5
=16/5