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prove sec2x = (sec^2x*csc^2x)/ (csc^2x - sec^2x)

+5
447
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prove

sec2x = (sec^2x*csc^2x)/ (csc^2x - sec^2x)
I tried to make it with no sucsess
Many thanks

Guest Feb 24, 2017

#2
+18956
+25

prove
sec2x = (sec^2x*csc^2x)/ (csc^2x - sec^2x)

$$\begin{array}{|rcll|} \hline && \frac{\sec^2(x)\cdot \csc^2(x)}{\csc^2(x) - \sec^2(x)} \\ &=& \frac{1}{ \frac{\csc^2(x) - \sec^2(x)}{\sec^2(x)\cdot \csc^2(x)} } \\ &=& \frac{1}{ \frac{\csc^2(x)}{\sec^2(x)\cdot \csc^2(x)} -\frac{\sec^2(x)}{\sec^2(x)\cdot \csc^2(x)} } \\ &=& \frac{1}{ \frac{1}{\sec^2(x)} -\frac{1}{ \csc^2(x)} } \quad & | \quad \frac{1}{\sec(x)}=\cos(x) \quad \frac{1}{ \csc(x)}=\sin(x) \\ &=& \frac{1}{ \cos^2(x) - \sin^2(x) } \quad & | \quad \cos(2x) = \cos^2(x) - \sin^2(x) \\ &=& \frac{1}{ \cos(2x) } \quad & | \quad \frac{1}{\cos(x)}=\sec(x)\\ &=& \sec(2x) \\ \hline \end{array}$$

heureka  Feb 24, 2017
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#1
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0

sec2x = (sec^2x*csc^2x)/ (csc^2x - sec^2x)    simplify the right side

(sec^2x*csc^2x)/ (csc^2x - sec^2x)   =

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

Get a common denominator for the fractions in the denominator  = sin^2xcos^2x

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

Invert the fraction in the denominator and multiply by the numerator

[ 1/(cos^2x sin^2x) ] * (sin^2x cos^2x)  / [cos^2x - sin^2x]  =

1 / [ cos^2x  - sin^2x ] =

1/ cos2x    =

sec2x

CPhill  Feb 24, 2017
#2
+18956
+25

prove
sec2x = (sec^2x*csc^2x)/ (csc^2x - sec^2x)

$$\begin{array}{|rcll|} \hline && \frac{\sec^2(x)\cdot \csc^2(x)}{\csc^2(x) - \sec^2(x)} \\ &=& \frac{1}{ \frac{\csc^2(x) - \sec^2(x)}{\sec^2(x)\cdot \csc^2(x)} } \\ &=& \frac{1}{ \frac{\csc^2(x)}{\sec^2(x)\cdot \csc^2(x)} -\frac{\sec^2(x)}{\sec^2(x)\cdot \csc^2(x)} } \\ &=& \frac{1}{ \frac{1}{\sec^2(x)} -\frac{1}{ \csc^2(x)} } \quad & | \quad \frac{1}{\sec(x)}=\cos(x) \quad \frac{1}{ \csc(x)}=\sin(x) \\ &=& \frac{1}{ \cos^2(x) - \sin^2(x) } \quad & | \quad \cos(2x) = \cos^2(x) - \sin^2(x) \\ &=& \frac{1}{ \cos(2x) } \quad & | \quad \frac{1}{\cos(x)}=\sec(x)\\ &=& \sec(2x) \\ \hline \end{array}$$

heureka  Feb 24, 2017

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