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Abbie23

  • 3 years ago

I just confused myself trying this : 1/sin^2(x) + Sec^2(x)/Tan^2(x)

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  1. lgbasallote
    • 3 years ago
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    change everything in terms of sines and cosines \[\sec^2 x = \frac{1}{\cos^2 x}\] \[\tan^2 x = \frac{\sin^2 x}{\cos^2 x}\]

  2. lgbasallote
    • 3 years ago
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    \[\large \frac{\sec^2 x}{\tan ^2 x} = \frac{\frac{1}{\cos^2 x}}{\frac{\sin^2 x}{\cos^2 x}}\] does that help?

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