例1(1)(2021·全国甲卷)若 alpha in (0,dfrac (pi )(2)), tan2α-|||-=dfrac (cos alpha )(2-sin alpha ) 则tanα等于 ()-|||-

若α∈（0，(π)/(2)），tan2α=(cosα)/(2-sinα)，则tanα=（ ）A. $\frac{\sqrt{15}}{15}$B. $\frac

7.[福建2020适应性模拟]已知 tan alpha =2 ,则-|||-dfrac ({sin )^2alpha -(cos )^2alpha +2}(2{

[题目]若 sin alpha =dfrac (1)(3), 则 cos 2alpha = ()()-|||-

设alpha =(dfrac (1)(2),0,0,dfrac (1)(2)) ,=E-(alpha )^Talpha =E+2(alpha )^Talpha

4.已知 sin alpha =-dfrac (3)(5), 且α是第三象限的角,则 cos alpha = __ ,-|||-tan alpha = __ o

(B) alpha lt dfrac (1)(2). (C) alpha geqslant dfrac (1)(2), (D) alpha gt dfrac (

1.“sin^2 alpha +sin^2 beta =1”是“sin alpha +cos beta =0”的（）.A. 充分非必要条件B. 必要非充分条件C

(sin x)=dfrac (1)({cos )^2x} in (0,dfrac (pi )(2)),则(sin x)=dfrac (1)({cos )^2x}

dfrac (Fx)(sin alpha ) D. dfrac (Fx)(cos alpha )

判断题-|||-tan (2pi +alpha )=tan alpha .-|||-39.-|||-A 对-|||-B 错