verify the identity cot(x-pi/2)=-tan x?

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  • 2 years ago
    Favorite Answer

    LHS=

    cos(x-pi/2)/sin(x-pi/2)=

    [cos(x)*0+sin(x)*1]/

    [sin(x)*0-cos(x)*1]=

    sin(x)/[-cos(x)]=

    -tan(x)=

    RHS.

  • cidyah
    Lv 7
    2 years ago

    cot(x-pi/2) = cos(x-pi/2) /sin(x-pi/2)

    cos(a-b) = cos(a)cos(b) + sin(a)sin(b)

    sin(a-b) = sin(a)cos(b) - cos(a)sin(b)

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

    = cos(x) (0) + sin(x) (1)

    = sin(x)

    cos(x-pi/2) = sin(x)

    sin(x-pi/2) = sin(x) cos(pi/2) - cos(x) sin(pi/2)

    = sin(x) (0) - cos(x) (1)

    = -cos(x)

    sin(x-pi/2) = -cos(x)

    cot(x-pi/2) = cos(x-pi/2)/sin(x-pi/2) = sin(x)/(-cos(x)) = -tan(x)

  • ted s
    Lv 7
    2 years ago

    depends upon what YOU know....cos ( x - π / 2 ) = sin x and sin (x - π / 2 ) = - cos x...thus their ratio is - tan x

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