Question

化简sin^2α*tanα+cos^2α*cotα+2sinα*cosα-cotα？

Answer 1

原式=sin²α*sin α/cosα+cos²α*cosα/sinα+2sinαcosα-cosα/sinα
通分
=(sin^4α+cos^4α+2sin²²αcos²α-cos²α)/sinαcosα
=[(sin²α+cos²α)²-cos²α]/sinαcosα
=(1-cos²α)/sinαcosα
=sin²α/sinαcosα
=sinα/cosα
=tanα,7,sin^2α*tanα+cos^2α*cotα+2sinα*cosα-cotα
=sin^3α/cosα+cos^3α/sinα+(2*sin^2α*cos^2α)/(sinα*cosα)-cotα
=(sin^4α+2*sin^2α*cos^2α+cos^4α)/(sinα*cosα)-cos^2α/(sinα*cosα)
=(1-cos^2α)/(sinα*cosα)
=sin^2α/(sinα*cosα)
=tanα,2,化简sin^2α*tanα+cos^2α*cotα+2sinα*cosα-cotα
详细解答,谢谢
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