meili 发表于 2022-10-27 15:38:05

已知(sina)^2÷(cosβ)^2+(cosa)^2(cosγ)^2=1,求证(tana)^2÷(tan)^2=(sinγ)^2设cosβ=t(t≥0),秋sinβ和tanβ

<p>问题:已知(sina)^2÷(cosβ)^2+(cosa)^2(cosγ)^2=1,求证(tana)^2÷(tan)^2=(sinγ)^2设cosβ=t(t≥0),秋sinβ和tanβ
<p>答案:↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">吉翔华的回答:<div class="content-b">网友采纳  (sina)^2÷(cosβ)^2+(cosa)^2(cosγ)^2=1(sina)^2÷(cosβ)^2+(cosa)^2(cosγ)^2=(sina)^2+(cosa)^2tana)^2÷(cosβ)^2+(cosγ)^2=tana)^2+1(tana)^2÷(cosβ)^2-(tana)^2=1-(cosγ)^2=(sinγ)^2(tana)^2*(1-(cos...
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查看完整版本: 已知(sina)^2÷(cosβ)^2+(cosa)^2(cosγ)^2=1,求证(tana)^2÷(tan)^2=(sinγ)^2设cosβ=t(t≥0),秋sinβ和tanβ