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(1-tan^4θ)cos^2θ+tan^2θ
人气:224 ℃ 时间:2020-06-09 18:18:19
解答
因为
(1-tan^4x)cos^2x+tan^2x
=(1+tan^2x)(1-tan^2x)cos^2x+tan^2x
=(1-tan^2x)[(cos^2x+sin^2x)/cos^2x]cos^2x+tan^2x
=1-tan^2x+tan^2x
=1
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