How can I simplify (1/cos2θ) - (1/cot2θ) using trigonometric identities?

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To simplify (1/cos2θ) - (1/cot2θ), start by expressing cot2θ in terms of sine and cosine, which is cot2θ = cos2θ/sin2θ. This allows the expression to be rewritten as (1/cos2θ) - (sin2θ/cos2θ). Combining the fractions yields (1 - sin2θ) / cos2θ. Recognizing that 1 - sin2θ equals cos2θ, the expression simplifies to cos2θ / cos2θ, resulting in 1. The final simplified expression is 1.
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Homework Statement



Simplify the following:

(1/cos2θ) - (1/cot2θ)

Homework Equations



Various trig identities

The Attempt at a Solution



I tried to make cos2θ into 1-sin2θ and cot2θ into csc2θ-1 but still couldn't find any obvious solution. Help?
 
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What do you get if you express cot as something using sines and cosines?? What if you then add up the fractions?
 
Did you try simply inverting both fractions? Does that show similarity to a trigonometric identity that you know?
 
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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