求证-2sin αcos α+1/1-2cos*2α=tan α-1/tan α+1

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求证-2sin αcos α+1/1-2cos*2α=tan α-1/tan α+1
求证-2sin αcos α+1/1-2cos*2α=tan α-1/tan α+1

求证-2sin αcos α+1/1-2cos*2α=tan α-1/tan α+1
证:(-2sinαcosα+1)/(1-2cos²α)=(sinα-cosα)²/(sin²α-cos²α)
=(sinα-cosα)²/[(sinα+cosα)(sinα-cosα)]
=(sinα-cosα)/(sinα+cosα)
=(tanα-1)/(tanα+1)