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Today, 13:10

Prove the trigonometric identity:

(csc^2x) / (cotx) = cscxsecx ... ?

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  1. Today, 13:23
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    Cscx = 1 / sinx, so csc^2x = 1 / sin²x, and cotx=cosx / sinx

    so (csc^2x) / (cotx) = (1 / sin²x) (sinx/cosx) = 1 / (sinxcosx) = 1/sinx 1/cosx

    finally

    (csc^2x) / (cotx) = cscxsecx
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