Results 1 to 2 of 2
Like Tree1Thanks
  • 1 Post By topsquark

Thread: Why does tan^2 y = x

  1. #1
    Member Furyan's Avatar
    Joined
    Sep 2011
    Posts
    132
    Thanks
    3

    Why does tan^2 y = x^2

    Hello,

    I continue to struggle with trig.

    The question is:

    Given that $\displaystyle x = \tan y$ show that $\displaystyle \dfrac{dy}{dx} = \dfrac{1}{1 + x^2}$.

    The answer given, most of which I follow, is:

    $\displaystyle \dfrac{d}{dy} (\tan y) = \sec^2 y$

    Taking the reciprocal of both sides.

    $\displaystyle \dfrac{dy}{dx} = \dfrac{1}{sec^2 y}$

    $\displaystyle \dfrac{dy}{dx} = \dfrac{1}{1 + \tan^2 y}$

    Hence, and this is the bit I don't get.

    $\displaystyle \dfrac{dy}{dx} = \dfrac{1}{1 + x^2}$

    It's bound to be something I'm supposed to know, something basic, but why is $\displaystyle \tan^2 y = x^2$

    Thanks
    Last edited by Furyan; Aug 28th 2012 at 04:33 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,152
    Thanks
    731
    Awards
    1

    Re: Why does tan^2 y = x

    Quote Originally Posted by Furyan View Post
    Hello,

    I continue to struggle with trig.

    The question is:

    Given that $\displaystyle x = \tan y$ show that $\displaystyle \dfrac{dy}{dx} = \dfrac{1}{1 + x^2}$.

    The answer given, most of which I follow, is:

    $\displaystyle \dfrac{d}{dy} (\tan y) = \sec^2 y$

    Taking the reciprocal of both sides.

    $\displaystyle \dfrac{dy}{dx} = \dfrac{1}{sec^2 y}$

    $\displaystyle \dfrac{dy}{dx} = \dfrac{1}{1 + \tan^2 y}$
    You have been given $\displaystyle x = tan(y)$. Then by squaring both sides: $\displaystyle tan^2(y) = x^2$.

    -Dan
    Thanks from Furyan
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum