# Proving Trig Identity again

• Dec 9th 2009, 08:08 AM
aceband
Proving Trig Identity again
Prove:

$\displaystyle (cot(x) + cosec(x))^2 = \frac{1+cos(x)}{1-cos(x)}$

And i can get this far but no further:

$\displaystyle \frac{cos^{2}(x) + 2cos(x) + 1}{1 - cos^{2}(x)}$

Anyone see my fault? Mybe i need to factorise the top of the factor...hmmmm
• Dec 9th 2009, 08:19 AM
alexmahone
Quote:

Originally Posted by aceband
Prove:

$\displaystyle (cot(x) + cosec(x))^2 = \frac{1+cos(x)}{1-cos(x)}$

And i can get this far but no further:

$\displaystyle \frac{cos^{2}(x) + 2cos(x) + 1}{1 - cos^{2}(x)}$

Anyone see my fault? Mybe i need to factorise the top of the factor...hmmmm

$\displaystyle (cotx+cosecx)^2=(\frac{cosx}{sinx}+\frac{1}{sinx}) ^2$

= $\displaystyle (\frac{cosx+1}{sinx})^2$

= $\displaystyle \frac{(cosx+1)^2}{sin^2x}$

= $\displaystyle \frac{(1+cosx)^2}{1-cosx^2}$

= $\displaystyle \frac{(1+cosx)^2}{(1+cosx)(1-cosx)}$

= $\displaystyle \frac{1+cosx}{1-cosx}$
• Dec 9th 2009, 08:33 AM
aceband
Really sorry but can't see how to get that to the final stage :S
• Dec 9th 2009, 08:35 AM
aceband
Ignore that comment i see it thanks