1.cosπ/3-tan9π/4+3/4tan²（-2π+π/6)+sin25π/6-cos²13π

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1.cosπ/3-tan9π/4+3/4tan²（-2π+π/6)+sin25π/6-cos²13π/6+sin17π/2
2.a,b,c为正数3的a次方=4的b次方=6的c次方证：2/a+1/b=2/c 3.sinx+cos+x=1/5(0＜x＜π）求tanx

1. 原式=cosπ/3-tanπ/4+3/4tan²π/6)+sinπ/6-cos²π/6+sinπ/2=1/2-1+1/4+1/2-3/2+1=-1/4
2. 原等式取ln函数,则,
aln3=bln4=cln6 令 aln3=bln4=cln6=t, 有 a=t/ln3 b=t/ln4 c=t/ln6
所以, 2/a+1/b=2ln3/t+ln4/t=(2ln3+ln4)//t=2ln6/t=2/c
3. 原等式两边平方,得 sin²x+2sinxcosx+cos²x=1/25
得 sin2x= -24/25=2sinxcosx/(sin²x+cos²x)
得 12tan²x+25tanx+12=0
得 tanx=-4/3或-3/4
又 0＜x＜π , tanx0 , cosx0 │sinx│>│cosx│ 即 │tanx│>1
故 tanx=-4/3
希望能帮到你.

求值sin25π/6+cos25π/3+tan(-25π/4)+sin(-7π/3)×cos(-13π/6)-sin(- sin25π/6+cos10π/3+tan(-25π/4)+sin(-7π/3)×cos(-13π/6)= 求sin25π/6+cos5π/3+tan(-25π/4)+sin(-7π/3)×cos(-13π/6)-sin(-5π cosπ/3+tanπ/4+3tan²π/6+sinπ/2+cosπ+sin3π/2等于?. 计算：sin25π/6+cos25π/4+tan(-25π/4)+sin(-26π/3) sin25π6 tan9π/8 sin25π/4+cos25π/3+tan(-25π/4)= 已知tanα/2=2,1.求tan(α+π/4)的值 2.（6sinα+cosα)/（3sinα-2cosα)的值 cosπ/3－tanπ/4+cosπ－sin3π/2 化简计算已知tanα/2=2求tan(α+π/4)的值(6sinα+cosα)/(3sinα-2cosα)的值 cos(19π/6)+tan(16π/3)=?