I'll abbreviate [tex]s=\sin\theta[/tex] and [tex]c=\cos\theta[/tex], so the identity to prove is
[tex]\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1s\right)[/tex]
On the left side, we can simplify a bit:
[tex]\dfrac{s+c+1}{s+c-1}=\dfrac{s+c-1+2}{s+c-1}=1+\dfrac2{s+c-1}[/tex]
[tex]\dfrac{1+s-c}{1-s+c}=-\dfrac{-2+1-s+c}{1-s+c}=-1+\dfrac2{1-s+c}[/tex]
Then
[tex]\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1{s+c-1}-\dfrac1{1-s+c}\right)[/tex]
So the establish the original equality, we need to show that
[tex]\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac1s[/tex]
Combine the fractions:
[tex]\dfrac{(1-s+c)-(s+c-1)}{(s+c-1)(1-s+c)}=\dfrac{2-2s}{c^2-s^2+2s-1}[/tex]
We can rewrite the denominator as
[tex]c^2-s^2+2s-1=c^2+s^2-2s^2+2s-1[/tex]
then using the fact that [tex]c^2+s^2=\cos^2\theta+\sin^2\theta=1[/tex], we get
[tex]1-2s^2+2s-1=2s-2s^2[/tex]
so that we have
[tex]\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac{2-2s}{2s-2s^2}=\dfrac1s[/tex]
as desired.
