How Do You Solve Trig Identities Using Basic Trigonometric Equations?

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The discussion focuses on solving the trigonometric identity (1 + cosθ) / (1 - cosθ) = (1 + secθ) / (secθ - 1) using basic trigonometric identities. Participants emphasize the importance of recognizing that cosθ is the reciprocal of secθ, which simplifies the problem significantly. By manipulating the right side of the equation with reciprocal identities, the solution is reached efficiently. The conversation highlights the challenges of learning trigonometric identities and the value of practice and guidance from teachers. Overall, mastering these identities requires time and effort, but understanding the relationships between functions can make the process easier.
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Homework Statement



(1 + cosθ) / (1 - cosθ) = (1 + secθ) / (secθ - 1)

Homework Equations



using only the quotient identities, pythagorean identities, and reciprocal identities

The Attempt at a Solution



didnt know where to start...
 
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lol Mark44.

Key word is reciprocal identity, and this should be obvious since cos\theta is the reciprocal of sec\theta and vice versa.
 
kk got it.

RS
=(1 + secθ) / (secθ - 1)
=(1 + 1/cosθ) / (1/cosθ - 1)
=([cosθ+1] / cosθ) / ([1-cosθ] / cosθ)
=[(cosθ+1)(cosθ)] / [(1-cosθ)(cosθ)]
cancel cosθ
=(1 + cosθ) / (1 - cosθ)
 
oh wow, all these question are so easy now, thanks for the advice about the reciprocal.
 
No problem :smile:

I remember starting off with trig identities... I would have to say that was the hardest and most stressful topic to learn. If you don't have someone like your teacher to help you out with those, you can get lost in a question for hours! Well, at least that's what me and a few of my friends did.

It all comes to you eventually with much practice on this topic though. Good luck.
 
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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