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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Prove : $\large{ \tt{ \frac{1}{ \sec \alpha + 1 } - \frac{ \cos\alpha }{ { \sin^{2} \alpha } } = \frac{ \cos \alpha }{ { \sin}^{2} \alpha } - \frac{1}{ \sec \alpha - 1} }}$
*Irrelevant / Random answers will be reported!