若tan(x-π)=2,求2sin^2x+1/cos^2x-sin^2x的值
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/09 16:19:46
若tan(x-π)=2,求2sin^2x+1/cos^2x-sin^2x的值
若tan(x-π)=2,求2sin^2x+1/cos^2x-sin^2x的值
若tan(x-π)=2,求2sin^2x+1/cos^2x-sin^2x的值
tan(x-π)=tanx=2
2sin²x+1/cos²x-sin²x
=2sin²x+sin²x+cos²x/cosx²x-sin²x
=3sin²x+cos²x/cos²x-sin²x
=[(3sin²x/cos²x)+(cos²x/cos²x)]/[(cos²x/cos²x)-(sin²x/cos²x)]
=(3tan²x+1)/(1-tan²x)=-13/3
2sin²x+1/cos²x-sin²x
=2sin²x+sin²x+cos²x/cosx²x-sin²x
=3sin²x+cos²x/cos²x-sin²x
=[(3sin²x/cos²x)+(cos²x/cos²x)]/[(cos²x/cos²x)-(sin²x/cos²x)]
=(3tan²x+1)/(1-tan²x)=-13/3
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