Question

各位数学高手帮忙解决下这几题吧~~tan(-15π+π/6)正切值

1.tan(-15π+π/6)正切值 2.已知(α属于π/2,π)，sinα=3/5,则tan(α+π/2)= 3.求函数y=tan(4x+π/6)的定义域，并指出它的单调区间 4.若α,β属于(π,3π/2),且tan^2α>tan^2β，则( ) A.α<β B.α>β C.α+β>3π D.α+β<2π . 在线等啊~~谢谢亲咯~~~~

Answer 1

1.tan(-15π+π/6)
=tan(π/6)
=√3/3
2.已知(α属于π/2,π)，sinα=3/5,则cosα=-4/5
tan(α+π/2)
=-tanα
=sinα/cosα
=3/4
3.求函数y=tan(4x+π/6)的定义域，并指出它的单调区间
4x+π/6不等于π/2+kπ
x属于R且x不等于π/12+kπ/4
-π/2+kπ<4x+π/6<π/2+kπ为增
-π/6+kπ/4<x<π/12+kπ/4
π/2+kπ<4x+π/6<3π/2+kπ为减
π/12+kπ/4<x<π/3+kπ/4
4.若α,β属于(π,3π/2),且tan^2α>tan^2β，则(
B
)
A.α<β
B.α>β
C.α+β>3π
D.α+β<2π
.