# If $tan^{-1} \bigg(\large \frac{x-1}{x-2} \bigg) $$+ \tan^{-1} \bigg(\large \frac{x+1}{x+2} \bigg) =\large \frac{\pi}{4}, find the value of x. ## 1 Answer Toolbox: • $$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 Mar 1, 2013
edited Mar 18, 2013