Browse Questions

# 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 $x$

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

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 $a+b = \large\frac{x-1}{x-2}+\large\frac{x+1}{x+2} = \large\frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)} = \large\frac{x^2+2x-x-2+x^2-2x+x-2}{(x-2)(x+2)} = \large\frac{2x^2-4}{(x-2)(x+2)}$