1. ## Prove the identity

Hi All,
one over sec theta + tan theta = sec theta - tan theta = cos theta over one + sin theta
1 / secѳ + tanѳ = sec ѳ – tanѳ = cosѳ / 1 + sinѳ
My job is to prove the identities
Any help would be appreciated

2. Originally Posted by Joel
Hi All,
one over sec theta + tan theta = sec theta - tan theta = cos theta over one + sin theta
1 / secѳ + tanѳ = sec ѳ – tanѳ = cosѳ / 1 + sinѳ
My job is to prove the identities
Any help would be appreciated
Please either use LaTeX or brackets, so that we know what you are talking about...

Is the first one $\frac{1}{\sec{\theta} + \tan{\theta}}$ or $\frac{1}{\sec{\theta}} + \tan{\theta}$?

3. Sorry its the first one

4. Originally Posted by Joel
Hi All,
one over sec theta + tan theta = sec theta - tan theta = cos theta over one + sin theta
1 / secѳ + tanѳ = sec ѳ – tanѳ = cosѳ / 1 + sinѳ
My job is to prove the identities
Any help would be appreciated

Second Identity:

$\frac{1}{\sec{\theta} + \tan{\theta}} = \frac{1}{\frac{1}{\cos{\theta}} + \frac{\sin{\theta}}{\cos{\theta}}}$

$= \frac{1}{\frac{1 + \sin{\theta}}{\cos{\theta}}}$

$= \frac{\cos{\theta}}{1 + \sin{\theta}}$

First identity:

$= \frac{\cos{\theta}(1 - \sin{\theta})}{1 - \sin^2{\theta}}$

$= \frac{\cos{\theta} - \cos{\theta}\sin{\theta}}{\cos^2{\theta}}$

$= \frac{1}{\cos{\theta}} - \frac{\sin{\theta}}{\cos{\theta}}$

$= \sec{\theta} - \tan{\theta}$.

5. Thanks heaps, your proofs are very neatly set out and clear.

Thankyou.