+0

-1
163
3
+13

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

Sep 23, 2019

#1
0

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

Sep 23, 2019
#2
+108650
+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
+24376
+1

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

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

Sep 24, 2019