Verifying Trigonometric Identities

AI Thread Summary
Verifying trigonometric identities can be challenging, especially when algebra becomes complex. The original problem presented was miswritten, leading to confusion in simplification. Correcting the expression to (cot(x) - tan(x)) / (sin(x)cos(x)) = csc^2(x) - sec^2(x) allows for clearer analysis. It is suggested to split the left-hand side into two fractions to simplify further. Addressing typos and clarifying expressions is crucial for effective problem-solving in trigonometry.
misssue
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I have been having lots of trouble verifying trigonometric identities. I know the fundamental identities but I am actually having trouble with the algebra that goes along with the problems.
The problem I am working on now is:

cos(x)-tan(x)/sin(x)cos(x) = csc^2 (x) - sec^2 (x)

(The csc and sec are squared. I didn't know the best way to right that on here)

I tried to change everything to sin/cos but I felt like I made the equation much more confusing doing that.

I got:

(cos(x)/sin(x))-(sin(x)/cos(x))/(sin(x)cos(x)) = (1/sin(x))^2 - (1/cos(x))^2

I am more than a little lost.

I also tried it by changing only the left side and got:

(cot(x)-tan(x)) (1/sin(x)) (1/cos(x))

With either option I don't know where to go next and I'm not even sure if I started correctly.
 
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misssue said:
I have been having lots of trouble verifying trigonometric identities. I know the fundamental identities but I am actually having trouble with the algebra that goes along with the problems.
The problem I am working on now is:

cos(x)-tan(x)/sin(x)cos(x) = csc^2 (x) - sec^2 (x)

(The csc and sec are squared. I didn't know the best way to right that on here)
Looks like you have a typo there, because as written the LHS won't simplify to the RHS. I think you mean this:
\frac{\cot x - \tan x}{\sin x \cos x} = \csc^2 x - \sec^2 x

(Also, "right" should be "write".)

In which case, why don't you start by splitting the LHS into a difference of 2 fractions and go from there.
 
I feel silly for writing "write" as "right". There was a typo. Thank you for figuring that out and for the tip!
 
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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