trig identities

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December 3rd 2012, 06:15 PM
ahmedb

trig identities

Hello, I am trying to prove this identity: (cscx-cotx)^2=(1-cosx)/(1+cosx)
currently I am did these steps:

(csc(x) - cot(x)) (csc(x) - cot(x)) = (cscx-cotx)^2=(1-cosx)/(1+cosx)
csc^2(x) - 2cot(x)csc(x) + cot^2(x) = (cscx-cotx)^2=(1-cosx)/(1+cosx)
(1/sin^2(x)) - 2(cos(x)/sin(x))(1/sin(x)) + (cos^2(x) / sin^2(x)) = (cscx-cotx)^2=(1-cosx)/(1+cosx)
(1 - 2cos(x) + cos^2(x)) / sin^2(x) = (cscx-cotx)^2=(1-cosx)/(1+cosx)

Now I am stuck :S
please help and thanks!
December 3rd 2012, 06:26 PM
Prove It

Re: trig identities

$\displaystyle \begin{align*} \frac{1 - \cos{x}}{1 + \cos{x}} &= \frac{\left( 1 - \cos{x} \right)^2}{\left( 1 + \cos{x} \right) \left( 1 - \cos{x} \right)} \\ &= \frac{\left( 1 - \cos{x} \right)^2}{1 - \cos^2{x}} \\ &= \frac{\left( 1 - \cos{x} \right)^2}{\sin^2{x}} \\ &= \left(\frac{1 - \cos{x}}{\sin{x}}\right)^2 \\ &= \left( \frac{1}{\sin{x}} - \frac{\cos{x}}{\sin{x}} \right)^2 \\ &= \left( \csc{x} - \cot{x} \right)^2 \end{align*}$
December 3rd 2012, 06:37 PM
ahmedb

Re: trig identities

Quote:

Originally Posted by Prove It

$\displaystyle \begin{align*} \frac{1 - \cos{x}}{1 + \cos{x}} &= \frac{\left( 1 - \cos{x} \right)^2}{\left( 1 + \cos{x} \right) \left( 1 - \cos{x} \right)} \\ &= \frac{\left( 1 - \cos{x} \right)^2}{1 - \cos^2{x}} \\ &= \frac{\left( 1 - \cos{x} \right)^2}{\sin^2{x}} \\ &= \left(\frac{1 - \cos{x}}{\sin{x}}\right)^2 \\ &= \left( \frac{1}{\sin{x}} - \frac{\cos{x}}{\sin{x}} \right)^2 \\ &= \left( \csc{x} - \cot{x} \right)^2 \end{align*}$

umm, is it not possible to continue from what I am doing?
btw, how did you know you had to multiply by (1-cosx)?
like I would never have thought of that!