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krypton

  • 4 years ago

d/dx (tanx-1)/(secx)

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  1. dumbcow
    • 4 years ago
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    \[\frac{\tan x -1}{\sec x} = \frac{\tan x}{\sec x} - \frac{1}{\sec x} = \sin x -\cos x\] \[(\sin x - \cos x)' = \cos x +\sin x\]

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