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=

若asecα-2tanα=1,bsecα+tanα=2则a^2+b^2=
asecα-2tanα=1 即a/cosα-2sinα/cosα=1 得a=cosα+2sinα
bsecα+tanα=2 即b/cosα+sinα/cosα=2 得b=2cosα-sinα
a^2+b^2=(cosα+2sinα )^2+(2cosα-sinα )^2
=(cosα)^2+4sinα*cosα+4(sinα)^2+4(cosα)^2-4sinα*cosα+(sinα)^2
=5[(cosα)^2+(sinα)^2]
=5