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ilya1733
@ilya1733
July 2021
1
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A) tg²α - sin²α - tg²α · sin²α
B) cos⁴x(1+tg²x)+sin²x
упростите выражение
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kirichekov
Verified answer
Tg²α-sin²α-tg²α*sin²α=tg²α(1-sin²α)-sin²α=tg²α*cos²α-sin²α=sin²α-sin²α=0
cos⁴x(1+tg²x)+sin²x=cos⁴x*(1/cos²x)+sin²x=cos²x+sin²x=1
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Answers & Comments
Verified answer
Tg²α-sin²α-tg²α*sin²α=tg²α(1-sin²α)-sin²α=tg²α*cos²α-sin²α=sin²α-sin²α=0cos⁴x(1+tg²x)+sin²x=cos⁴x*(1/cos²x)+sin²x=cos²x+sin²x=1