bentuk sederhana ....

[tex] \frac{sin \: \alpha }{1 + cos \: \alpha } + \frac{1 + cos \: \alpha }{ \sin\alpha } = \frac{ {sin}^{2} \alpha }{(1 + cos \: \alpha )sin \: \alpha } + \frac{1 + 2cos \: \alpha + {cos}^{2} \alpha }{(1 + cos \: \alpha )sin \: } \alpha \\ = \frac{ {sin}^{2} \alpha + {cos}^{2} \alpha + 1 + 2cos \: \alpha }{ (1 + cos \: \alpha )sin \: \alpha} \\ = \frac{1 + 1 + 2cos \: \alpha }{(1 + cos \: \alpha )sin \: \alpha} \\ = \frac{2 + 2cos \: \alpha }{(1 + cos \: \alpha )sin \: \alpha} \\ = \frac{2(1 + cos \: \alpha )}{(1 + cos \: \alpha )sin \: \alpha} \\ = \frac{2}{sin \: \alpha } \\ = 2cosec \: \alpha [/tex]
Jadi jawabanny E. 2 cosec alfa