已知tan=根号3,求cosa-sina的值?
2个回答
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原式
=根号下[(1+sin
α)*(1+sin
α)/(1-sin
α)(1+sin
α)]-根号下[(1-sin
α)*(1-sin
α)/(1+sin
α)(1-sin
α)]
=根号下[(1+sinα)²/(1-sin²α)]-根号下[(1-sinα)²/(1-sin²α)]
=根号下[(1+sinα)²/cos²α]-根号下[(1-sinα)²/cos²α]
=-(1+sin
α)/cosα+(1-sin
α)/cos
α(α为第二象限角,去掉根号时注意变号)
=-2sinα/cosα
=-2tanα
=根号下[(1+sin
α)*(1+sin
α)/(1-sin
α)(1+sin
α)]-根号下[(1-sin
α)*(1-sin
α)/(1+sin
α)(1-sin
α)]
=根号下[(1+sinα)²/(1-sin²α)]-根号下[(1-sinα)²/(1-sin²α)]
=根号下[(1+sinα)²/cos²α]-根号下[(1-sinα)²/cos²α]
=-(1+sin
α)/cosα+(1-sin
α)/cos
α(α为第二象限角,去掉根号时注意变号)
=-2sinα/cosα
=-2tanα
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