(Ⅰ)求cosα的值;
(Ⅱ)求的值.
解:(Ⅰ)∵∴π<α-
∴sin(α-)=-
∴cosα=cos[(α-)+]
=cos(α-)cos--sin(α-)sin
=
(Ⅱ)原式=2sinαcosαtan(-α)
=2
=-2×