College Trig Question?

How do I verify the identity cotx/cscx - cscx/cotx = -sinxtanx ?

3 Answers

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  • 4 weeks ago

    cotx => cosx/sinx and cosecx = 1/sinx

    so, cotx/cosecx = (cosx/sinx)/(1/sinx) = cosx

    Hence, cosecx/cotx => 1/cosx

    Then, cotx/cosecx - cosecx/cotx = cosx - (1/cosx)

    so, RHS = (cos²x - 1)/cosx

    Now, as sin²x + cos²x = 1, then cos²x - 1 = -sin²x

    Then, (cos²x - 1)/cosx => -sin²x/cosx

    i.e. -sinx(sinx/cosx) => -sinxtanx

    :)>    

  • 4 weeks ago

    cotx/cscx - cscx/cotx =

    cotx * sinx - cscx * tanx =(cosx/sinx) * sinx - (1/sinx) * sinx/cosx =

    cosx - 1/cosx =

    cos^2x/cosx - 1/cosx =

    (cos^2x - 1)/cosx =

    -(1 - cos^2x)/cosx =

    -sin^2x/cosx =

    -sinx sinx/cosx =

    -sinx tanx

  • 4 weeks ago

    cot(x)/csc(x) - csc(x)/cot(x)

    = cos(x) - sec(x)

    = -sin(x) tan(x)

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