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关于三角函数的证明题,证明:1+sinα/cosα=tanα+secα-1/tanα-secα+1
题目内容:
关于三角函数的证明题,
证明:
1+sinα/cosα =tanα+secα-1/tanα-secα+1优质解答
右边=[(sina/cosa)+(1/cosa)-1]/[(sina/cosa)-(1/cosa)+1] 分子分母同乘以cosa,得:
右边=[sina+1-cosa]/[sina-1+cosa]
=[2sin(a/2)cos(a/2)+1-(1-2sin²(a/2))]/[2sin(a/2)cos(a/2)-1+(1-2sin²(a/2))]
=[2sin(a/2)cos(a/2)+2sin²(a/2)]/[2sin(a/2)cos(a/2)-2sin²(a/2)]
=[cos(a/2)+sin(a/2)]/[cos(a/2)-sin(a/2)]
=[cos(a/2)+sin(a/2)]²/[cos²(a/2)-sin²(a/2)]
=(1+sina)/cosa=左边.
证明:
1+sinα/cosα =tanα+secα-1/tanα-secα+1
优质解答
右边=[sina+1-cosa]/[sina-1+cosa]
=[2sin(a/2)cos(a/2)+1-(1-2sin²(a/2))]/[2sin(a/2)cos(a/2)-1+(1-2sin²(a/2))]
=[2sin(a/2)cos(a/2)+2sin²(a/2)]/[2sin(a/2)cos(a/2)-2sin²(a/2)]
=[cos(a/2)+sin(a/2)]/[cos(a/2)-sin(a/2)]
=[cos(a/2)+sin(a/2)]²/[cos²(a/2)-sin²(a/2)]
=(1+sina)/cosa=左边.
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