Multiple Choice Question (MCQ)

2\tan^{-1}\dfrac{1}{3} +\tan^{-1}\dfrac{1}{7} =

  1. ×

    \dfrac{11}{49}

  2. ×

    \dfrac{\pi}{6}

  3. \dfrac{\pi}{4}

  4. ×

    \dfrac{\pi}{2}

Collected in the board: Inverse function

Steven Zheng posted 1 month ago


Answer


  1. Let

    \arctan(1/3)=m<\dfrac{\pi}{4} ,

    \arctan(1/7)=n<\dfrac{\pi}{4}

    We get,

    \tan m =\dfrac{1}{3}, \tan n=\dfrac{1}{7}

    \tan 2m=\dfrac{2\tan m}{1-\tan^2 m}=\dfrac{2\cdotp \dfrac{1}{3} }{1-(\dfrac{1}{3})^2 } =\dfrac{3}{4}

    \tan(2m+n)=\dfrac{\tan 2m+\tan n}{1-\tan 2m\cdotp \tan n } =\dfrac{\dfrac{3}{4}+\dfrac{1}{7} }{1-\dfrac{3}{4}\cdotp \dfrac{1}{7} } =1


    Therefore, 2m+n=\dfrac{\pi}{4}

    2\arctan \dfrac{1}{3} +\arctan\dfrac{1}{7} =\dfrac{\pi}{4}


Steven Zheng posted 1 month ago

Scroll to Top