已知a为锐角,且tana=2,\\\\\\\\\\\\\已知a为锐角,且tana=2,求sin a+cos a/sin a-cos a的值/boob_zy ,为啥 sin a-cos a 会是1/2 呢,,

sin a+cos a/sin a-cos a (说明：分子分母同时处以cos a即可)
=tan a+1/tana-1
=3/1
=3

(sin a+cos a)/(sin a-cos a)
=[(sin a+cos a)/cos a]/[(sin a-cos a)/cos a]
=[tana+1]/[tana-1]
=[2+1]/[2-1]
=3

是(sin a+cos a)/(sin a-cos a)还是sin a+cos a/sin a-cos a无括号?
是(sin a+cos a)/(sin a-cos a)的话
=[(sin a+cos a)/cos a]/[(sin a-cos a)/cos a]
=[tana+1]/[tana-1]
=[2+1]/[2-1]
=3 (tan a=sin a/cos a)
是sin a+cos a/sin a-cos a无括号的话
tan a=2,故sin a=2cos a
代入（sin a）^2+(cos a)^2=1得cos a=根号5/5
sin a+cos a/sin a-cos a
=cos a+1/2
=根号5/5+1/2
cos a/sin a=1/tan a=1/2,sin a-cos a=cos a

(sin a+cos a)/(sin a-cos a),分子,分母同时乘以1/cosa,得到
(tana+1)/(tana-1)=(2+1)/(2-1)=3.