Question

已知sin^2α/sin^2β+cos^2αcos^2γ=1,求证：tan^2α/tan^2β=sin^2γ，非常感谢！

Answer 1

证明:sin^2α/sin^2β+cos^2αcos^2γ=1,
变形:sin^2α/sin^2β+cos^2α[1-sin^2γ]=1,
即: sin^2α/sin^2β+cos^2α-cos^2αsin^2γ=1,
解出:cos^2αsin^2γ=sin^2α/sin^2β+cos^2α-1,
整理:cos^2αsin^2γ=sin^2α/sin^2β-sin^2α.
再整理:cos^2αsin^2γ=sin^2α[1/sin^2β-1].
再变形:cos^2αsin^2γ=sin^2α[(1-sin^2β)/sin^2β].
即:cos^2αsin^2γ=sin^2α[(cos^2β)/sin^2β]
即:cos^2αsin^2γ=sin^2α[cot^2β)]
两端除以:cos^2α,
得:sin^2γ=tan^2α[cot^2β]
即:sin^2γ=(tan^2α)/tan^2β
或:(tan^2α)/tan^2β=sin^2γ
证明毕.