Mathematics
Mathematics, 18.05.2021 15:50, rebeccamckellpidge

Prove : \large{ \tt{ \frac{1}{ \sec \alpha + 1 } -  \frac{ \cos\alpha  }{ { \sin^{2}  \alpha } }   =  \frac{ \cos \alpha }{ { \sin}^{2}  \alpha }  -  \frac{1}{ \sec \alpha  - 1} }}
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