Question

求详解：若tana=2，则（sin³a＋cosa）／（sin³a＋sina）的值为？

Answer 1

（sin³a＋cosa）／（sin³a＋sina）(上下同除以cosa)

=(sin^2tana+1)/[(sin^2a+1)*tana]
=(2sin^2a+1)/(2sin^2a+2)
=(2sin^2a+sin^2a+cos^2a)/(2sin^2a+2sin^2a+2cos^2a)
=(3sin^2a+cos^2a)/(4sin^2a+2cos^2a) (上下同除以cos^2a)
=(2tan^2a+1)/(4tan^2a+2)
=9/18
=1/2

Answer 2

sin/cosx=2
cosx=sinx/2
代入sin²a+cos²a=1
sin²a=4/5

上下除以cosx
原式=(sin²*sina/cosa+1)/(sin²a*sina/cosa+sina/cosa)
=(sin²atana+1)/(sin²atana+tana)
=自己算一下吧

Answer 3

原式上下除以cos a
原
=(tan a*sin^2a+1)/(tan a*sin^2a+tana)
可以算出sina =2/根号5
代入=13/18