cos²A/1+sin²A-cos²A/1-sin²A=.........?
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cos²A/1+sin²A-cos²A/1-sin²A=.........?
1 Jawaban
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1. Jawaban alex422
[tex] \frac{ {cos}^{2} \alpha}{1 + {sin}^{2} \alpha } - \frac{ {cos}^{2} \alpha }{1 - {sin}^{2} \alpha } = \frac{ {cos}^{2} \alpha }{1 + {sin}^{2} \alpha } - (\frac{1 - {sin}^{2} \alpha }{1 - {sin}^{2} \alpha } ) = \\ \frac{ {cos}^{2} \alpha - (1 + {sin}^{2} \alpha ) }{1 + {sin}^{2} \alpha } = \frac{ - ( {sin}^{2} \alpha - {cos}^{2} \alpha + {sin}^{2} \alpha + {cos}^{2} \alpha ) }{1 + {sin}^{2} \alpha } \\ = \frac{ -2 {sin}^{2} \alpha }{1 + {sin}^{2} \alpha } [/tex]