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Lv 7
? asked in Science & MathematicsMathematics · 9 years ago

differentiate tan inverse (cotx) + cot inverse(tanx)?

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  • Erika Russell
    Lv 4
    9 years ago
    Favorite Answer

    y = arctan(cotx) + arccot(tanx)

    dy/dx = [-csc^2(x)] / [1 + cot^2(x)] - sec^2(x) / [ 1 + tan^2(x)]

    simplify a bit

    1 + cot^2(x) = csc^2(x)

    1 + tan^2(x) = sec^2(x)

    dy/dx = -csc^2(x) / csc^2(x) - sec^2(x) / sec^2(x)

    dy/dx = -1 - 1

    dy/dx = -2

  • ?
    Lv 4
    9 years ago

    let f(x)=tan inverse cot(x) +cot inverse tan(x)

    =(pi/2 -x)+(pi/2 -x)

    =pi -2x

    f'(x) =-2

    Source(s): tan(x)= cot(pi/2-x)
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