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BIGBANG11
@BIGBANG11
July 2022
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Докажите равенство:
а) 1+tg^2a=1/cos^2a
б) 1/(cos^2a) -1=tg^2a
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Answers & Comments
MizoriesKun
а) 1+tg^2a=1/cos^2a
1 + (sin²a/сos²a)=(cos²a+sin²a)/cos²a =1/cos²a
б) 1/(cos^2a) -1=tg^2a
1/(cos²a)-1= ((sin²a+cos²a) -cos²a) /cjs²a = sin²a/cos²a =tg²a
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Answers & Comments
1 + (sin²a/сos²a)=(cos²a+sin²a)/cos²a =1/cos²a
б) 1/(cos^2a) -1=tg^2a
1/(cos²a)-1= ((sin²a+cos²a) -cos²a) /cjs²a = sin²a/cos²a =tg²a