Proving the Identity: sinA/(1+cosA)=tan(A/2)

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The discussion centers on proving the trigonometric identity sinA / (1+cosA) = tan(A/2). Participants express confusion about using the double angle formula for cos2ø to solve the problem. One user indicates they are struggling to apply the formulas correctly. Ultimately, the original poster states that the problem has been resolved and no further replies are needed. The conversation highlights the challenges some users face when working with trigonometric identities.
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Prove the identity:

sinA / (1+cosA) = tan (A/2).

Im going no where with this, please help?:confused: :confused:
 
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use the double angle formula for cos2ø.
 
Sorry I don't understand. And I am trying to apply the double angle formulae but I am just getting confused.
 
Problem solved...No need to reply
 
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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