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Rauau1999
@Rauau1999
July 2022
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ТРИГОНОМЕТРИЯ. Доказать тождество:
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oganesbagoyan
Verified answer
√(1+sinα)/(1-sinα) - √(1+sinα)/(1-sinα) = -2tqα ,90°<α<180°.
----
√(1+sinα)/(1-sinα) - √(1-sinα)/(1+sinα) =
√(1+sinα)²/(1-sinα)((1+sinα) - √(1-sinα)²/(1+sinα)(1-sinα) =
√(1+sinα)²/(1-sin²α) - √(1-sinα)²/(1-sin²α) =
√(1+sinα)²/cos²α - √(1-sinα)²/cos²α =
(1+sinα)/|cosα| - (1-sinα)/|cosα| =(1+sinα - 1+sinα)/|cos
α| =
2sinα/|cosα| = 2sinα/(-cosα) = -2tq
α ч.н.д.
* * * но 90°<α<180°⇒cosα<0 поэтому |cosα|= -cosα * * *
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Answers & Comments
Verified answer
√(1+sinα)/(1-sinα) - √(1+sinα)/(1-sinα) = -2tqα ,90°<α<180°.----
√(1+sinα)/(1-sinα) - √(1-sinα)/(1+sinα) =
√(1+sinα)²/(1-sinα)((1+sinα) - √(1-sinα)²/(1+sinα)(1-sinα) =
√(1+sinα)²/(1-sin²α) - √(1-sinα)²/(1-sin²α) =
√(1+sinα)²/cos²α - √(1-sinα)²/cos²α =
(1+sinα)/|cosα| - (1-sinα)/|cosα| =(1+sinα - 1+sinα)/|cosα| =
2sinα/|cosα| = 2sinα/(-cosα) = -2tqα ч.н.д.
* * * но 90°<α<180°⇒cosα<0 поэтому |cosα|= -cosα * * *