若1+sinα1−sinα-1−sinα1+sinα=-2tanα,则角α的取值范围是_.
问题描述:
若
-
1+sinα 1−sinα
=-2tanα,则角α的取值范围是______.
1−sinα 1+sinα
答
已知等式变形得:(1+sinα)(1−sinα)(1−sinα)2-(1−sinα)(1+sinα)(1+sinα)2=|cosα|1−sinα-|cosα|1+sinα=|cosα|(1+sinα−1+sinα)cos2α=2sinα|cosα|cos2α=2sinα|cosα|=-2tanα=−2sinαcosα,∴|...