Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 years ago

prove cos(x) - sec(x) = -tan(x) sin(x)?

4 Answers

Relevance
  • 2 years ago
    Favorite Answer

    secx = 1/cosx, so considering the LHS we have:

    cosx - (1/cosx) => (cos²x - 1)/cosx

    Now, sin²x + cos²x = 1

    so, cos²x - 1 = -sin²x

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

    or, (-sinx/cosx)(sinx)

    => -tanxsinx = RHS

    :)>

  • cidyah
    Lv 7
    2 years ago

    cos(x)-sec(x)

    =cos(x) -1/cos(x)

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

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

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

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

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

    = -tan (x) sin (x)

  • 2 years ago

    LHS= cos(x) - [ 1/cos(x) ]

    =[ cos²(x) - 1 ] / cos(x)

    = - [1 - cos²(x) ]/ cos (x)

    = - sin²(x)/ cos(x)

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

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

    = - tan(x)sin(x)=RHS

    Learn :

    •secx = 1/cos x

    • 1 - cos²x = sin²x

    • sinx / cos x = tanx

  • 2 years ago

    cos(x) - sec(x) (LHS)

    = cos(x) - (1/cos(x))

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

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

    = (-sin(x)/cos(x)) * sin(x)

    = -tan(x)sin(x) (RHS). Done.

Still have questions? Get your answers by asking now.