Question

已知tanα=2/3,求[(cosα-sinα)/(cosα+sinα)]+[(cosα+sinα)/(cosα-sinα)]的值

Answer 1

[(cosα-sinα)/(cosα+sinα)]+[(cosα+sinα)/(cosα-sinα)]
=[(cosα/cosα-sinα/cosα)/(cosα/cosα+sinα/cosα)]+[(cosα/cosα+sinα/cosα)/(cosα/cosα-sinα/cosα)]
=[(1-tanα)/(1+tanα)]+[(1+tanα)/(1-tanα)]
=[(1-2/3)/(1+2/3)]+[(1+2/3)/(1-2/3)]
=(1/3)/(5/3)+(5/3)/(1/3)
=1/5+5
=26/5

Answer 2

[(cosα-sinα)/(cosα+sinα)]+[(cosα+sinα)/(cosα-sinα)]
分子分母同除以cosa
=[(1-tana)/(1+tana)]+[(1+tana)/(1-tana)]
=[(1/3)/(5/3)]+[(5/3)/(1/3)]
=1/5+5
=26/5