tanα+tanα=? tanα-tanα=? 2tanα=? (1+tanα)/(1-tanα)=? (1-tanα)/(1+tanα)=?一定要准确啊,考试要用到了
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![tanα+tanα=? tanα-tanα=? 2tanα=? (1+tanα)/(1-tanα)=? (1-tanα)/(1+tanα)=?一定要准确啊,考试要用到了](/uploads/image/z/5193345-57-5.jpg?t=tan%CE%B1%2Btan%CE%B1%3D%3F+tan%CE%B1-tan%CE%B1%3D%3F+2tan%CE%B1%3D%3F+%EF%BC%881%2Btan%CE%B1%EF%BC%89%2F%281-tan%CE%B1%29%3D%3F+%EF%BC%881-tan%CE%B1%EF%BC%89%2F%281%2Btan%CE%B1%29%3D%3F%E4%B8%80%E5%AE%9A%E8%A6%81%E5%87%86%E7%A1%AE%E5%95%8A%2C%E8%80%83%E8%AF%95%E8%A6%81%E7%94%A8%E5%88%B0%E4%BA%86)
tanα+tanα=? tanα-tanα=? 2tanα=? (1+tanα)/(1-tanα)=? (1-tanα)/(1+tanα)=?一定要准确啊,考试要用到了
tanα+tanα=? tanα-tanα=? 2tanα=? (1+tanα)/(1-tanα)=? (1-tanα)/(1+tanα)=?
一定要准确啊,考试要用到了
tanα+tanα=? tanα-tanα=? 2tanα=? (1+tanα)/(1-tanα)=? (1-tanα)/(1+tanα)=?一定要准确啊,考试要用到了
tanα+tanα = 2tanα = 2(sinα)/(cosα)
tanα-tanα = 0
2tanα = 2(sinα)/(cosα)
(1+tanα)/(1-tanα) = (tan45°+tanα)/(1-tan45°tanα) = tan(α+45°)
(1-tanα)/(1+tanα) = (tan45°-tanα)/(1+tan45°tanα) = tan(45°-α) = -tan(α-45°)
证明:tanα+tanβ=tan(α+β)-tanαtanβtan(α+β)
tanα
tanα+1/tanα
证明tanα+tanβ=tan(α+β)-tanαtanβtan(α+β)
tanα+tanβ=tan(α+β)-tanαtanβtan(α+β)证明
证明tanα+tanβ=tan(α+β)-tanαtanβtan(α+β)
tanα+tanβ+tan(α+β)tanαtanβ=tan(α+β)求证左边等于右边
证明 tanα=2tanβ
化简tanα+tan(45-α)+tanαtan(45-α)=?请写下过程,
2tanα/(1-tanαtanα)=-4/3求tanα
求证:(tanα+tanβ)/(tanα-tanβ)=sin(α+β)/sin(α-β)
2tanα/(1-tanαtanα)=-4/3求tanα要过程
已知tanα+tanβ=2,tan(α+β)=4,tanα
α+β=45º,则tanα+tanβ+tanαtanβ
tan
tan
tan
已知tanα+tanβ=2,tan(α+β)=4 则tanα×tanβ=