已知sinα=asinβ,tanα=btanβ,α为锐角,求证:(cosα)^2=(a^2-1)/(b^2-1).
<p>问题：已知sinα=asinβ,tanα=btanβ,α为锐角,求证:(cosα)^2=(a^2-1)/(b^2-1).<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">何佩琨的回答：<div class="content-b">网友采纳 (sinβ)^2=(sinα)^2/a^2,(cosβ)^2=1-(sinβ)^2=/a^2. (tanβ)^2=(sinα)^2/. (tanα)^2=b^2*(tanβ)^2=b^2(sinα)^2/. 1/(cosα)^2=b^2/. b^2(cosα)^2=a^2-1+(cosα)^2. (cosα)^2=(a^2-1)/(b^2-1).
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