作业帮若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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