How do you prove this using trigonometic identities...

1 + sin theta / 1 - sin theta = 2 tan squared theta + 1 + 2 tan theta secant theta

You mean

\frac{1+sin( \theta)}{1-sin( \theta)} = 2tan^2 \theta + 2tan( \theta )sec( \theta )+1

?

\frac{1+sin{\theta}}{1-sin{\theta}}=2tan^{2}{\theta}+2tan{\theta}sec{\the ta}+1


Multiply the left side by \frac{1+sin{\theta}}{1+sin{\theta}}:

\frac{1+sin{\theta}}{1-sin{\theta}}\cdot\frac{1+sin{\theta}}{1+sin{\theta }}

Now, with a little wrangling and noting that 1-sin^{2}{\theta}=cos^{2}{\theta}, you should be able to hammer it into shape. Write back if you remain stuck.