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\(\frac{sec²(4/x)-1}{cos(6/x)-cos(2/x)}\)

 Sep 23, 2019
 #1
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0

What do you want done with it? Solve for x? Expand it? Or what?

 Sep 23, 2019
 #2
avatar+105634 
+2

\(\frac{sec²(\frac{x}{4})-1}{cos(\frac{6}{x})-cos(\frac{2}{x})}\\~\\ =(sec²(\frac{x}{4})-1)\cdot \frac{1}{cos(\frac{6}{x})-cos(\frac{2}{x})}\\~\\ =\frac{sin^2(\frac{4}{x})}{Cos^2(\frac{4}{x})}\cdot \frac{1}{cos(\frac{4}{x}+\frac{2}{x})-cos(\frac{2}{x})}\\~\\ \)

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\(cos(\frac{4}{x}+\frac{2}{x})-cos(\frac{2}{x})\\~\\ =cos(\frac{2}{x})cos(\frac{4}{x})-sin(\frac{2}{x})sin(\frac{4}{x})-cos(\frac{2}{x})\\~\\ =cos(\frac{2}{x})[cos^2(\frac {2}{x})-sin^2(\frac{2}{x})]-sin(\frac{2}{x})[2sin(\frac{2}{x})cos(\frac{2}{x})]-cos(\frac{x}{2})\\~\\ =cos(\frac{2}{x})\{cos^2(\frac {2}{x})-sin^2(\frac{2}{x})-2sin^2(\frac{2}{x})-1\}\\~\\ =cos(\frac{2}{x})\{-sin^2(\frac {2}{x})-sin^2(\frac{2}{x})-2sin^2(\frac{2}{x})\}\\~\\ =cos(\frac{2}{x})\{-4sin^2(\frac{2}{x})\}\\~\\ =-4cos(\frac{2}{x})sin^2(\frac{2}{x})\\~\\ \)

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\(\frac{sin^2(\frac{4}{x})}{Cos^2(\frac{4}{x})}\\~\\ =\frac{(sin(\frac{4}{x}))^2}{(Cos(\frac{4}{x}))^2}\\~\\ =\frac{(2sin(\frac{2}{x})cos(\frac{2}{x}))^2}{(Cos^2(\frac{2}{x})-sin^2(\frac{2}{x}))^2}\\~\\ =\frac{4sin^2(\frac{2}{x})cos^2(\frac{2}{x})}{Cos^2(\frac{4}{x})}\\~\\ \)

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\(=\frac{4sin^2(\frac{2}{x})cos^2(\frac{2}{x})}{Cos^2(\frac{4}{x})}\cdot\frac{1}{-4cos(\frac{2}{x})sin^2(\frac{2}{x})}\\~\\ =\frac{-cos(\frac{2}{x})}{Cos^2(\frac{4}{x})}\\~\\ =-cos(\frac{2}{x})Sec^2(\frac{4}{x})\\~\\\)

 

\(\boxed{=-cos\left(\frac{2}{x}\right)Sec^2\left(\frac{4}{x}\right)}\)

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 Sep 23, 2019
 #3
avatar+23317 
+1

​ Please help

\(\dfrac{\sec^2 \left(\dfrac{4}{x} \right)-1}{\cos \left(\dfrac{6}{x} \right)-\cos \left(\dfrac{2}{x} \right)} \)


\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{\sec^2 \left(\dfrac{4}{x} \right)-1}{\cos \left(\dfrac{6}{x} \right)-\cos \left(\dfrac{2}{x} \right)}} \\\\ &&\qquad \boxed{\sec^2 \left(\dfrac{4}{x} \right)-1 = \dfrac{\sin^2 \left(\dfrac{4}{x} \right)}{\cos^2 \left(\dfrac{4}{x} \right)} \\~\\= \sin^2 \left(\dfrac{4}{x} \right)\sec^2 \left(\dfrac{4}{x} \right) } \\\\ &=& \dfrac{\sin^2 \left(\dfrac{4}{x} \right)\sec^2 \left(\dfrac{4}{x} \right) } {\cos \left(\dfrac{6}{x} \right)-\cos \left(\dfrac{2}{x} \right)} \\\\ &&\qquad \boxed{ \cos \left(\dfrac{6}{x} \right)-\cos \left(\dfrac{2}{x} \right) = -2\sin \left( \dfrac{6+2}{x} \above 1pt 2 \right) \sin \left( \dfrac{6-2}{x}\above 1pt 2 \right) \\~\\= -2\sin \left(\dfrac{4}{x} \right) \sin \left(\dfrac{2}{x} \right) } \\\\ &=& \dfrac{\sin^2 \left(\dfrac{4}{x} \right)\sec^2 \left(\dfrac{4}{x} \right) } {-2\sin \left(\dfrac{4}{x} \right) \sin \left(\dfrac{2}{x} \right)} \\\\ &=& -\dfrac{\sin \left(\dfrac{4}{x} \right)\sec^2 \left(\dfrac{4}{x} \right) } {2 \sin \left(\dfrac{2}{x} \right)} \\\\ &=& -\dfrac{\sin \left(2\cdot \dfrac{2}{x} \right)\sec^2 \left(\dfrac{4}{x} \right) } {2 \sin \left(\dfrac{2}{x} \right)} \\\\ &&\qquad \boxed{ \sin \left(2\cdot \dfrac{2}{x} \right) = 2 \sin \left(\dfrac{2}{x} \right)\cos \left(\dfrac{2}{x} \right) } \\\\ &=& -\dfrac{ 2 \sin \left(\dfrac{2}{x} \right)\cos \left(\dfrac{2}{x} \right)\sec^2 \left(\dfrac{4}{x} \right) } {2 \sin \left(\dfrac{2}{x} \right)} \\\\ &=& \mathbf{-\cos \left(\dfrac{2}{x} \right)\sec^2 \left(\dfrac{4}{x} \right)} \\ \hline \end{array}\)

 

 

laugh

 Sep 24, 2019

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