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.======以下答案可供参考======供参考答案1:asecα -2tanα =1asecα=2tanα +1(asecα)^2=(2tanα +1)^2a^2*(secα)^2=(2tanα +1)^2a^2*(secα)^2=4(tanα)^2+4tanα +1a^2*[1+(tanα)^2]=4(tanα)^2+4tanα +1.................1bsecα +tanα =2bsecα =2-tanα (bsecα )^2=(2-tanα)^2 b^2*(secα )^2=(tanα)^2 -4tanα+4b^2*[1+(tanα)^2]=(tanα)^2 -4tanα+4.....................21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

这个解释是对的

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