Question

solve the problem step by step

Answer 1

$= > 1 + \frac{sec \alpha }{sec \alpha } = \frac{ {sin}^{2} \alpha }{1 - cos \alpha } \\ \\ = > 1 + 1 = \frac{ 1 - {cos}^{2} \alpha }{1 - cos \alpha } \\ \\ = > 2 = \frac{(1 - cos \alpha )(1 + cos \alpha) }{(1 - cos \alpha) } \\ \\ = > 2 = 1 + cos \alpha \\ \\ = > cos \alpha = 2 - 1 = 1$
then...

$= > \alpha = {cos}^{ - 1} 1 \\ \\ = > \alpha = 0 \: \: degree$
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