Prove this identity.

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

AdamTaurus May 1, 2019

#1**+1 **

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

CPhill May 1, 2019

#3**+2 **

**Prove this identity.**

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

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

heureka May 2, 2019