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Math Help - Inverse tan

  1. #1
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    Inverse tan

    Using the expansion of tan (a+b) find the value of k such that tan^{-1} k + tan^{-1} (\frac{3}{5}) = \frac{\pi}{4}
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  2. #2
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    Quote Originally Posted by nerdzor View Post
    Using the expansion of tan (a+b) find the value of k such that tan^{-1} k + tan^{-1} (\frac{3}{5}) = \frac{\pi}{4}
    Let \tan^{-1} k = a \Rightarrow \tan a = k and let \tan^{-1} \frac{3}{5} = b \Rightarrow \tan b = \frac{3}{5} and let [tex].

    Then a + b = \frac{\pi}{4} \Rightarrow \tan (a + b) = 1.

    Now substitute the expressions for \tan a and \tan b into the expansion of \tan (a + b) and solve for k.
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  3. #3
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    Hello, nerdzor!

    We can do this head-on ... if you dare.

    We need: . \tan (A + B) \:=\:\frac{\tan A + \tan B}{1 - \tan A\tan B}


    Using the expansion of \tan (A+B), find the value of k

    such that: . \tan^{-1}(k) + \tan^{-1}\!\left(\tfrac{3}{5}\right) \:=\:\frac{\pi}{4}

    Take the tangent of both sides: . \tan\bigg[\tan^{-1}(k) + \tan^{-1}\!\left(\tfrac{3}{5}\right)\bigg] \;=\;\tan\left(\tfrac{\pi}{4}\right)

    . . . . \frac{\tan\left[\tan^{-1}(k)\right] + \tan\left[\tan^{-1}(\frac{3}{5})\right]} {1 - \tan\left[\tan^{-1}(k)\right]\,\tan\left[\tan^{-1}(\frac{3}{5})\right]} \;=\;1 . \Longrightarrow\quad \frac{k + \frac{3}{5}}{1 - \frac{3}{5}k} \:=\:1

    . . . . k + \frac{3}{5} \;=\;1 - \frac{3}{5}k \quad\Longrightarrow\quad \frac{8}{5}k \;=\;\frac{2}{5} \quad\Longrightarrow\quad \boxed{k \:=\:\frac{1}{4}}

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