Results 1 to 4 of 4

Math Help - Finding tan(90-β)

  1. #1
    Newbie
    Joined
    Aug 2010
    Posts
    8

    Finding tan(90-β)

    I have trouble doing this problem.

    The problem:

    Let tan β= 6
    Find tan(90-β).

    Please and thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by luckystar001 View Post
    I have trouble doing this problem.

    The problem:

    Let tan β= 6
    Find tan(90-β).

    Please and thank you.
    \displaystyle \tan(90^\circ - \beta) = \cot{\beta} = \frac{1}{6}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008
    Just incase you need to show WHY \tan{(90^{\circ} - \beta)} = \cot{\beta}...


    \tan{(90^{\circ} - \beta)} = \frac{\sin{(90^{\circ} - \beta)}}{\cos{(90^{\circ} - \beta)}}

     = \frac{\cos{\beta}}{\sin{\beta}}

     = \cot{\beta}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by luckystar001 View Post
    I have trouble doing this problem.

    The problem:

    Let tan β= 6
    Find tan(90-β).

    Please and thank you.
    Hi luckystar001,

    Draw a right-angled triangle,

    label the two acute angles \beta and 90^o-\beta

    since the two acute angles must sum to 90^o

    as all three angles sum to 180^o

    From the definitions of \displaystyle\huge\ Sin\theta=\frac{opposite}{hypotenuse}

    \displaystyle\huge\ Cos\theta=\frac{adjacent}{hypotenuse}

    you can see the logic in the above posts.


    A much less appetizing way is as follows...

    \displaystyle\huge\tan\left(90^o-\beta\right)=\frac{tan90^0+tan(-\beta)}{1-tan90^otan(-\beta)}

    As \theta\rightarrow\ 90^o,\ tan\theta\rightarrow\ \infty

    hence we must take limits by dividing all terms by tan90^o

    In truth we are multiplying by 1=\displaystyle\huge\frac{\left(\frac{1}{tan90^o}\  right)}{\left(\frac{1}{tan90^o}\right)}

    \displaystyle\huge\lim_{\theta\rightarrow90^o}\lef  t(\frac{1+\frac{tan(-\beta)}{tan90^o}}{\frac{1}{tan90^o}-tan(-\beta)}\right)=\frac{1}{tan\beta}

    as tan(-\beta)=-tan\beta
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: December 8th 2011, 10:27 AM
  2. Replies: 1
    Last Post: July 3rd 2010, 10:40 PM
  3. Finding a limit. Finding Maclaurin series.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 18th 2010, 10:04 PM
  4. Finding the radius, solving, and finding theta?
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: June 13th 2009, 02:37 PM
  5. Replies: 1
    Last Post: April 9th 2009, 09:02 AM

Search Tags


/mathhelpforum @mathhelpforum