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If \( \tan^{-1} \frac{\large x-1}{\large x-2} + \tan^{-1} \frac{\large x+1}{\large x+2} = \frac{\large \pi}{\large 4} \) then find the value of \(x\)

$\begin{array}{1 1} \pm \frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2} \\ - \frac{1}{\sqrt 2} \\ \pm \sqrt 2 \end{array} $

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Toolbox:
  • \(tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy}\: xy < 1\)
  • tan\(\large\frac{\pi}{4}=1\)
Given $tan^{-1} \large\frac{\large x-1}{\large x-2} + tan^{-1} \large\frac{\large x+1}{\large x+2} = \large\frac{\large \pi}{\large 4}$:
We know that \(tan^{-1}a+tan^{-1}b=tan^{-1}\frac{a+b}{1-ab}\: ab < 1\)
 
Let us substitue for $a$ and $b$ such that $a = \large\frac{\large x-1}{\large x-2}, b = \large\frac{\large x+1}{\large x+2}$ in $tan^{-1} \large\frac{\large x-1}{\large x-2} + tan^{-1} \large\frac{\large x+1}{\large x+2}$:
 
Numerator $a+b = \large\frac{x-1}{x-2}+\large\frac{x+1}{x+2} = \large\frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)} = \large\frac{x^2+2x-x-2+x^2-2x+x-2}{(x-2)(x+2)} = \large\frac{2x^2-4}{(x-2)(x+2)}$
Denominator $1 - ab = 1- (\large\frac{x-1}{x-2})(\large\frac{x+1}{x+2}) = \large\frac{ (x-2)(x+2) - (x-1)(x+1)}{(x-2)(x+2)} =$$\large\frac{x^2 - 2x + 2x -4 -x^2+x-x+1}{(x-2)(x+2)} = \large\frac{-3}{(x-2)(x+2)}$
$\large \frac{a+b}{1-ab} = \large\frac{2x^2-4}{(x-2)(x+2)} \times \frac{(x-2)(x+2)}{-3} = \large\frac{2x^2-4}{-3}$
$\Rightarrow tan^{-1} \large\frac{\large x-1}{\large x-2} + tan^{-1} \large\frac{\large x+1}{\large x+2} = tan^{-1}\bigg(\large\frac{2x^2-4}{-3} \bigg)$
 
The above equation reduces to $ tan^{-1} \large\frac{\large x-1}{\large x-2} + tan^{-1} \large\frac{\large x+1}{\large x+2} =tan^{-1} \bigg(\large\frac{2x^2-4}{-3} \bigg) = \large\frac{\pi}{4}$
$ tan^{-1} \bigg(\large\frac{2x^2-4}{-3} \bigg) = \large\frac{\pi}{4} \rightarrow \bigg(\large\frac{2x^2-4}{-3} \bigg) = tan \large\frac{\pi}{4}$
 
We know that $tan \large\frac{\pi}{4}=1$.
 
\( \therefore\) $ \bigg(\large\frac{2x^2-4}{-3} \bigg) = 1 \rightarrow 2x^2 - 4 = -3 \rightarrow 2x^2 = 1$.
\( x = \pm \large\frac{1}{\sqrt 2}\)

 

answered Feb 22, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1
 

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