Prove

I'll use x instead of "theta"  [ easier to type.....LOL!!! ]

csc x + 1

_______    =

csc x -  1

1/sin x  + 1

_________  =

 1/sinx - 1

(1 + sin x ) / sinx

______________  =

(1 - sin x) / sin x

(1 + sin x)           sin x

________   *  ________  =

   sin x              1 - sin x

1 + sin x

________  =   the left hand side

1 - sin x

Prove this identity.

\(\large \dfrac{1+\sin(\theta)}{1-\sin(\theta)}=\dfrac{\csc(\theta)+1}{\csc(\theta)-1}\)

\(\begin{array}{|rcll|} \hline && \dfrac{1+\sin(\theta)}{1-\sin(\theta)} \\\\ &=& \dfrac{1+\sin(\theta)}{1-\sin(\theta)}\cdot \dfrac{ \dfrac{1}{\sin(\theta)} } {\dfrac{1}{\sin(\theta)}} \\\\ &=& \dfrac{ \Big(1+\sin(\theta)\Big) \dfrac{1}{\sin(\theta)} } {\Big(1-\sin(\theta) \Big)\dfrac{1}{\sin(\theta)}} \\\\ &=& \dfrac{ \dfrac{1}{\sin(\theta)} + \dfrac{\sin(\theta)} {\sin(\theta)} } { \dfrac{1}{\sin(\theta)} - \dfrac{\sin(\theta)}{\sin(\theta)} } \\\\ &=& \dfrac{ \dfrac{1}{\sin(\theta)} + 1 } { \dfrac{1}{\sin(\theta)} - 1 } \quad | \quad \boxed{\dfrac{1}{\sin(\theta)} = \csc(\theta) } \\\\ &=& \dfrac{ \csc(\theta) + 1 } { \csc(\theta) - 1 } \\ \hline \end{array}\)