Question

1 + sin theta by 1 minus sin theta is equal to secant theta + tan theta the whole square

Answer 1

$({ \sec(?) + \tan(?) })^{2} \\ (\frac{1}{ \cos(?) } + \frac{ \sin(?) }{ \cos(?) } ) ^{2} \\ { \frac{1 + \sin(?) }{ \cos(?) ^{2} } }^{2} \\ \frac{1 + { \sin(?) }^{2} }{1 - { \sin(?) }^{2} } \\ \frac{(1 + \sin(?))(1 + \sin(?) ) }{(1 + \sin(?))(1 - \sin(?)) } \\ \frac{1 + \sin(?) }{1 - \sin(?) }$
theta is replaced by ?


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