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  • 16-02-2017
  • Mathematics
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Sec^2x-1/sec^2x=sin^2x

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LammettHash
LammettHash LammettHash
  • 16-02-2017
[tex]\dfrac{\sec^2x-1}{\sec^2x}=1-\dfrac1{\sec^2x}=1-\cos^2x=\sin^2x[/tex]
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