若asecα -2tanα =1,bsecα +tanα =2则a^2+b^2的结果是

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若asecα -2tanα =1,bsecα +tanα =2则a^2+b^2的结果是
若asecα -2tanα =1,bsecα +tanα =2则a^2+b^2的结果是

若asecα -2tanα =1,bsecα +tanα =2则a^2+b^2的结果是
asecα -2tanα =1,
则a=(1+2tanα)/ secα
=(1+2tanα)cosα
= cosα+2sinα.
bsecα +tanα =2,
则b=(2- tanα)/ secα
=(2- tanα) cosα
=2 cosα- sinα.
则a^2+b^2=( cosα+2sinα)^2+(2 cosα- sinα)^2
= cosα^2+4 sinα^2++4 cosα^2+ sinα^2
=5.

asecα -2tanα =1
asecα=2tanα +1
(asecα)^2=(2tanα +1)^2
a^2*(secα)^2=(2tanα +1)^2
a^2*(secα)^2=4(tanα)^2+4tanα +1
a^2*[1+(tanα)^2]=4(tanα)^2+4tanα +1.................1
bsecα +tan...

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asecα -2tanα =1
asecα=2tanα +1
(asecα)^2=(2tanα +1)^2
a^2*(secα)^2=(2tanα +1)^2
a^2*(secα)^2=4(tanα)^2+4tanα +1
a^2*[1+(tanα)^2]=4(tanα)^2+4tanα +1.................1
bsecα +tanα =2
bsecα =2-tanα
(bsecα )^2=(2-tanα)^2
b^2*(secα )^2=(tanα)^2 -4tanα+4
b^2*[1+(tanα)^2]=(tanα)^2 -4tanα+4.....................2
1式+2式得
（a^2+b^2)*[1+(tanα)^2]=4(tanα)^2+4tanα +1+(tanα)^2 -4tanα+4
（a^2+b^2)*[1+(tanα)^2]=5(tanα)^2+5
（a^2+b^2)*[1+(tanα)^2]=5[(tanα)^2+1]
a^2+b^2=5

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若asecα-2tanα=1,bsecα+tanα=2则a^2+b^2= 若asecα-2tanα=1,bsecα+tanα=2则a^2+b^2= 若asecα -2tanα =1,bsecα +tanα =2则a^2+b^2的结果是若asec-2tan=1,bsec+tan=2,则a^2+b^2= 参数方程x=asecα,y=btanα(α是参数,-π/2 这个不定积分做的对不对dx/√(a^2+x^2) 令x=atant dx=d(atant)=asec^2tdt原式=∫asec^2tdt/√(a^2+a^2*(tant)^2=∫asec^2tdt/a√(1+tant^2=∫asec^2tdt/asect=∫sectdt=1/2ln|(1+sint)/(1-sint)|+c=ln|sect+tant|+c因为x=atant 所以tant=x/a s 2tanα／(1-tanαtanα)=-4／3求tanα 2tanα／(1-tanαtanα)=-4／3求tanα要过程 tanα+tanα=? tanα-tanα=? 2tanα=? （1+tanα）/(1-tanα)=? （1-tanα）/(1+tanα)=?一定要准确啊,考试要用到了证明tan(α)*tan(β)+tan(β)*tan(γ)+tan(α)*tan(γ)=1 (α+β+γ=π/2)详细一点证明tanαtanβ+tanαtanγ+tanβtanγ=1,α+β+γ=π/2 已知∠α+∠β+∠γ=π／2 求证tanαtanβ+tanαtanγ+tanβtanγ=1 若tanα=1/3,tanβ=-1/4,则tan(2α-2β)= tanα=1/2 tan方α等于多少若tanα=1/3,tan（β-α）=-2,则tanβ的值为化简1+tanαtanα/2 证明 tanα=2tanβ 若α、β为锐角,(1+tanα/2)(1+tanβ/2)=2,则tanαtanβ=