Precalculus: proving trigonometric identity

AI Thread Summary
The discussion centers on the attempt to prove the trigonometric identity tan(1 + cos(x))^2 = 1 - cos(x). Participants express confusion over the validity of the identity, with one user noting that substituting x=0 shows the left and right sides yield different results, indicating it is not an identity. The consensus is that the equation cannot be proven as a true identity due to inconsistencies when evaluating specific values of x. Additionally, it is suggested that the problem may be misphrased and is more about solving rather than proving. Ultimately, the conclusion is that the expression does not hold as a valid trigonometric identity.
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Homework Statement



prove that: tan(1+cos(x))^2 = 1-cos(x)

Homework Equations



trig identities, like the pythagorean, sum/difference, double/half angle identities, power reducing identities, etc...

The Attempt at a Solution



i'm not sure where to start; i tried using the pythagorean identity where 1+tan(X)^2 = sec(x)^2, but couldn't get anywhere after that :\

then i used my calculator and made X a random number. i typed in the left side expression, and pressed enter. i then typed in the right hand expression, and pressed enter. the two values were different. what did i do wrong?
i'm not really sure anymore that its even possible to prove the above...
 
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You're right. It's not an identity. Just put x=0. Then you don't even need a calculator. Must be some mistake here.
 
If I'm write you want to write tan(1 + cos x)^2 = 1 - cos(x). I think it is not possible to prove because if cos (x) is zero then {tan}^2{1} is not equal to 1. Even if the equation is tan(1 + {cos}^2{x}) then also you can't prove it. I think, the problem is not to prove but to solve.
 
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
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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