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harbi · May 31, 2007, 05:25 AM

How do you prove this using trigonometic identities...

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

Capuchin · May 31, 2007, 05:27 AM

You mean

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

?

galactus · May 31, 2007, 09:01 AM

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

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.