Prove problem involving sec cos sin

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The discussion revolves around proving the identity (sec(x) - cos(x)) / (sec(x) + cos(x)) = (sin^2(x)) / (1 + cos^2(x)). Participants suggest rewriting sec(x) as 1/cos(x) to simplify the problem. They recommend combining terms in both the numerator and denominator by finding a common denominator. The focus is on step-by-step guidance to ensure clarity in the solution process. Ultimately, the thread aims to help users understand the proof through detailed explanations.
MathRaven
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Hi guys been struggling to solve this
If you know how to do it, please do solve and show work so that I can follow your steps


Homework Statement



prove (sec(x)-cos(x))/(sec(x)+cos(x))=(sin^2(x))/(1+cos^2(x))
 
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sec(x) = 1/cos(x). So, write both secant terms in this way. Then, separately in both the numerator and the denominator, combine the two terms by finding their common denominator. What do you end up with?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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