Can you help me solve this trig identity question?

AI Thread Summary
The discussion revolves around solving a trigonometric identity involving cotangent and cosine functions. The user is struggling to simplify the expression and has made some progress but is confused about the next steps. They have converted the left side of the equation into sine and cosine terms, but clarity is needed on the subsequent transformation. The conversation highlights the importance of understanding trigonometric identities and their conversions for simplification. Assistance is requested to navigate through the complexities of the problem.
x.xmedzx.x
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hey guyz...ok iv been trying to figure this question out for so long...and i jus can't.i get up to a certain point and then i jus get confused.so if anyone can help me that would be great!

Prov that:
Cos^2x + Cotx ÷ Cos^2x – Cotx = Cos^2x (tanx) + 1 ÷ Cos^2x (tanx) -1
 
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\frac{\cos^{2}x + \cot x}{\cos^{2}x-\cot x} = \frac{\cos^{2}x(\tan x)+1}{\cos^{2} x(\tan x) -1}

Convert the left side to sines and cosines:

\frac{\cos^{2}x + \frac{\cos x}{\sin x}}{\cos^{2}x - \frac{\cos x}{\sin x}} = \frac{\cos^{2}x\sin x + \cos x}{\sin x}\frac{\sin x}{\cos^{2}x\sin x - \cos x} = \frac{\cos^{2}x\sin x + \cos x}{\cos^{2}x\sin x - \cos x} = \frac{\cos^{2}x(\sin x + \frac{1}{\cos x})}{\cos^{2}x(\sin x - \frac{1}{\cos x})}.

Can you go from there?
 
Last edited:
umm ok the first setp i got...but the 2nd one I am a little bit confused as to what you did...
 
x.xmedzx.x said:
umm ok the first setp i got...but the 2nd one I am a little bit confused as to what you did...
Note that;

\cot\theta=\frac{1}{\tan\theta}=\frac{1}{\frac{\sin\theta}{\cos\theta}}=\frac{\cos\theta}{\sin\theta}
 
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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