证明:cosθ−sinθ+1cosθ+sinθ−1=cosecθ+cotθ
待办事项
我们需要证明cosθ−sinθ+1cosθ+sinθ−1=cosecθ+cotθ.
解答
我们知道:
sin2A+cos2A=1
cosec2A−cot2A=1
sec2A−tan2A=1
cotA=cosAsinA
tanA=sinAcosA
cosecA=1sinA
secA=1cosA
因此,
cosθ−sinθ+1cosθ+sinθ−1=cosθsinθ−sinθsinθ+1sinθcosθsinθ+sinθsinθ−1sinθ (将每一项除以sinθ)
=cotθ−1+cosecθcotθ+1−cosecθ
=(cotθ+cosecθ−1)(cotθ−cosecθ+1)
=(cotθ+cosecθ−1)(cotθ+cosecθ)[(cotθ−cosecθ)+1](cotθ+cosecθ) (乘以和除以cotθ+cosecθ)
=(cotθ+cosecθ−1)(cotθ+cosecθ)(cotθ−cosecθ)(cotθ+cosecθ)+cotθ+cosecθ
=(cotθ+cosecθ−1)(cotθ+cosecθ)(cot2θ−cosec2θ)+cotθ+cosecθ
=(cotθ+cosecθ−1)(cotθ+cosecθ)−1+cotθ+cosecθ
=(cotθ+cosecθ−1)(cotθ+cosecθ)(cotθ+cosecθ−1)
=cotθ+cosecθ
=cosecθ+cotθ
证毕。
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