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# Solve the following for x : $tan^{-1} \bigg( \frac{x-1}{x-2} \bigg) + tan^{-1} \bigg( \frac{x+1}{x+2} \bigg) = \frac{\pi}{4}$

This question is Q.No.15 of sec, 2 chapter 2
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## 1 Answer

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

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