[tan(2π－α)·sin(2π－α)·cos(6π－α)·sin(3π/2 －α)]／[cos(－α)·sin(5π+α)·cos(3π/2 －α)·sin(α－π)]

若a属于(0,π/2)试比较tanα、tan(tanα)、tan(sinα) 若sinα＝3sin（π/2－α）,则tanα＋1/tanα－1＝ 设sin(α+π/2)=－1/4,且tanα＞0,求sinα,cosα和tanα的值 sin｛π－α｝cos｛2π－α｝tan｛－α＋π｝/sin｛π＋α｝化简 化简：[sin(﹣α)cos(π－α)]∕[tan(π＋α)sin﹙3π／2＋α)] 1.求证tanαsinα/tanα-sinα=tanα+sinα/tanαsinα2.已知sinα+cosα=-1/5(π/2 已知tan（π-α）=－2／1,则sinαcosα-2sin^2α= 化简cos2(-α)-tan(2π+α)/sin(-α) sin(2π-α)tan(α+3π)tan(-α-π)/tan(α-3π)cos(π-α) 化简cos(α -π )tan(α-2π)tan(2π-α)/sin(π+α) cos(π-α)tan(2π-α)tan(π-α)/sin(π+α) 化简 化简sin(2π-α）tan（α+π）tan（-α-π）/ Sinπ/3Tanπ/3+tanπ/6COSπ/6－Tanπ/4COSπ/2 求证tan²α－sin²α＝tan²α·sin²α 化简：(sin^2α-tan^2α)/(sin^2α·tan^2α) sin²(-α)·cos(π+α)/tan(2∏+α)tan(∏+α)cos³(-∏-α) 已知tanα-tanβ=2tanα^2tanβ,且αβ均不等于kπ/2.试求sinβsin(2α+β）/sinβ或者能给思路已知就是这个：tanα-tanβ=2tanαtanαtanβ [sin(2π+α）·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α）·tan(2π+α）化简