prove
sec2x = (sec^2x*csc^2x)/ (csc^2x - sec^2x)
I tried to make it with no sucsess
Many thanks
+26367 prove
sec2x = (sec^2x*csc^2x)/ (csc^2x - sec^2x)
\(\begin{array}{|rcll|} \hline && \frac{\sec^2(x)\cdot \csc^2(x)}{\csc^2(x) - \sec^2(x)} \\ &=& \frac{1}{ \frac{\csc^2(x) - \sec^2(x)}{\sec^2(x)\cdot \csc^2(x)} } \\ &=& \frac{1}{ \frac{\csc^2(x)}{\sec^2(x)\cdot \csc^2(x)} -\frac{\sec^2(x)}{\sec^2(x)\cdot \csc^2(x)} } \\ &=& \frac{1}{ \frac{1}{\sec^2(x)} -\frac{1}{ \csc^2(x)} } \quad & | \quad \frac{1}{\sec(x)}=\cos(x) \quad \frac{1}{ \csc(x)}=\sin(x) \\ &=& \frac{1}{ \cos^2(x) - \sin^2(x) } \quad & | \quad \cos(2x) = \cos^2(x) - \sin^2(x) \\ &=& \frac{1}{ \cos(2x) } \quad & | \quad \frac{1}{\cos(x)}=\sec(x)\\ &=& \sec(2x) \\ \hline \end{array} \)
+128460 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
+26367 prove
sec2x = (sec^2x*csc^2x)/ (csc^2x - sec^2x)
\(\begin{array}{|rcll|} \hline && \frac{\sec^2(x)\cdot \csc^2(x)}{\csc^2(x) - \sec^2(x)} \\ &=& \frac{1}{ \frac{\csc^2(x) - \sec^2(x)}{\sec^2(x)\cdot \csc^2(x)} } \\ &=& \frac{1}{ \frac{\csc^2(x)}{\sec^2(x)\cdot \csc^2(x)} -\frac{\sec^2(x)}{\sec^2(x)\cdot \csc^2(x)} } \\ &=& \frac{1}{ \frac{1}{\sec^2(x)} -\frac{1}{ \csc^2(x)} } \quad & | \quad \frac{1}{\sec(x)}=\cos(x) \quad \frac{1}{ \csc(x)}=\sin(x) \\ &=& \frac{1}{ \cos^2(x) - \sin^2(x) } \quad & | \quad \cos(2x) = \cos^2(x) - \sin^2(x) \\ &=& \frac{1}{ \cos(2x) } \quad & | \quad \frac{1}{\cos(x)}=\sec(x)\\ &=& \sec(2x) \\ \hline \end{array} \)
