Prove the identity

Joel · #1 $Old$ November 2nd, 2009, 07:13 PM

Hi All,
This reads as
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

Prove It · #2 $Old$ November 2nd, 2009, 07:17 PM

Quote:

Originally Posted by Joel $View Post$

Hi All,
This reads as
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 $\frac{1}{\sec{\theta}} + \tan{\theta}$ ?

Joel · #3 $Old$ November 2nd, 2009, 07:21 PM

Sorry its the first one

Prove It · #4 $Old$ November 2nd, 2009, 07:29 PM

Quote:

Originally Posted by Joel $View Post$

Hi All,
This reads as
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:

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

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

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

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

Joel · #5 $Old$ November 2nd, 2009, 07:45 PM

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

Thankyou.