Multiple Choice Question (MCQ)

If \sin x+\cos x = \dfrac{1}{5}, and 0 < x <\pi, then the value of \tan x

  1. -\dfrac{4}{3}

  2. ×

    -\dfrac{4}{3} or -\dfrac{3}{4}

  3. ×

    -\dfrac{3}{4}

  4. ×

    \dfrac{3}{4} or -\dfrac{4}{3}

Collected in the board: Trigonometry

Steven Zheng posted 2 months ago


Answer

  1. Square the both sides of the equation of

    \sin x+\cos x = \dfrac{1}{5}

    We get,

    \sin^2 x+\cos^2 x+2\sin x\cos x = \dfrac{1}{25}

    Apply Pythagorean Identity

    2\sin x\cos x = - \dfrac{24}{25}
    (1)

    Apply double angle identity for sine function,

    \sin 2x = - \dfrac{24}{25}

    \pi <2x<2\pi

    \dfrac{\pi}{2} < x< \pi
    (2)
    1- 2\sin x\cos x = 1-( - \dfrac{24}{25}) = \dfrac{49}{25}


    (\sin x-\cos x)^2 = \dfrac{49}{25}

    \sin x-\cos x = \dfrac{7}{5} (because \dfrac{\pi}{2} < x< \pi)

    Solve the equation system

    \sin x-\cos x = \dfrac{7}{5}

    \sin x+\cos x = \dfrac{1}{5}

    Wr get

    \sin x = \dfrac{4}{5}, \cos x = -\dfrac{3}{5}

    Therefore,

    \tan = \dfrac{\sin x}{\cos x} = -\dfrac{4}{3}


    A is the choice

Steven Zheng posted 2 months ago

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