zhuhb
幼苗
共回答了23个问题采纳率:91.3% 举报
1.=根号下{[(1+sinα)(1-sinα)]/[(1-sinα)(1-sinα)]}-根号下{[(1-sinα)(1+sinα)]/[(1+sinα)(1+sinα)]}
=根号下{(1-sinα*sinα)/([1-sinα)(1-sinα)]}-根号下{[(1-sinα*sinα)/[(1+sinα)(1+sinα)]}
=[cosα/(1-sinα)]-[cosα/(1+sinα)]
=cosα(1+sinα-1+sinα)/(1-sinα*sinα)
=2sinα/cosα
=2tanα
1年前
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