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SandraASA
@SandraASA
August 2022
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tg^2a-sin^2a/tg^2(п-a)sin^2(a-1260) помогите пожалуйста!!!
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yarovoe
Tg²α-sin²α/tg²(π-a)sin²(α-1260°)= (tg²α-sin²α)/tg²α·sin²α=1/sin²α-1/tg²α=
=1/sin²α-1/sin²α/cos²α=1/sin²α-cos²α/sin²α=(1-cos²α)/sin²α=sin²α/sin²α=1
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Answers & Comments
=1/sin²α-1/sin²α/cos²α=1/sin²α-cos²α/sin²α=(1-cos²α)/sin²α=sin²α/sin²α=1