2cos π/2-tan π/4+3/4tan的平方 π/6-sin π/6+cos的平方 π/6 +sin 3π/2 求计算过程及...2cos π/2-tan π/4+3/4tan的平方 π/6-sin π/6+cos的平方 π/6 +sin 3π/2

Sinπ/3Tanπ/3+tanπ/6COSπ/6－Tanπ/4COSπ/2 cosπ/3+tanπ/4+3tan²π/6+sinπ/2+cosπ+sin3π/2等于?. 化简:(sin(α-3π)*cos(-α))/(tan(α-5π)*cos^2(α-5π)*tan(4π-α)) cosπ/3－tanπ/4+cosπ－sin3π/2 化简计算 求值:2cosπ/2-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6+sin3π/2 -1/4cosπ 已知tan(π+α)=-1/3 tan(α+β)=[sin(π-2α)+4(cosα)^2]/[10(cosα)^2-sin2α]求……已知tan(π+α)=-1/3tan(α+β)=[sin(π-2α)+4(cosα)^2]/[10(cosα)^2-sin2α]求：1、tan(α+β)2、tanβ 化简cos(α-π)sin²(α+2π)/tan (α+4π)tan(α+π)cos³(-α-π) 求证2(cosθ -sinθ )/1+sinθ +cosθ =tan(π/4-θ /2)-tanθ /2 已知tan(π/4+α)=-1/2,求4cosα(sinα+cosα)/(1-tanα) 求证：tanα+1/tan(π/4+α/2)=1/cosα帮个忙! 求证：1/cosα-tanα=1/tan（π/4+α/2） 求证：tanα+1/tan(π/4+α/2)=1/cosα 过程清楚一点 计算（1）cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6[tan(-150°)cos（-210°）cos420°]/cot(-600°)sin（-1050°）cos25/6π+cos25/3π+tan(-25/4π)5sinπ/2+2cos0-3sin3/2π+10cosπt 计算cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6[tan(-150°)cos（-210°）cos420°]/cot(-600°)sin（-1050°）cos25/6π+cos25/3π+tan(-25/4π)5sinπ/2+2cos0-3sin3/2π+10cosπtan10°t 计算2cosπ/2-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6+sin3π/2同上 写出过程 2cosπ/2—tanπ/4+3/4tan平方π/6+cos平方π/6+sin3π/2 tan(a-4π)cos(π+a)sin^2(a+3π)/tan(3π+a)cos^2(2/5π+2)=0.5,求tana tan(a-4π)cos(π+a)sin^2(a+3π)/tan(3t+a)cos^2(2/5π+2)=-2/1求cosa*|tana|