Question

The value(s) of $x$ satisfying the equation ${\tan }^{-1}\frac{x-1}{x-2}+{\tan }^{-1}\frac{x+1}{x+2}=\frac{\pi }{4}$ is (are)

• A. $\sqrt{2}$
• B. $-\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{-1}{\sqrt{2}}$

Answer 1

Answer: (C), (D)

$\frac{-1}{\sqrt{2}}$

Let $A={\tan }^{-1}\frac{x-1}{x-2}$ and $B={\tan }^{-1}\frac{x+1}{x+2}$
Then, $\tan A=\frac{x-1}{x-2}$ and $\tan B=\frac{x+1}{x+2}$

Now, $A+B=\frac{\pi }{4}$
$\Rightarrow \tan (A+B)=\tan \frac{\pi }{4}$
$\Rightarrow \frac{\tan A+\tan B}{1-\tan A\tan B}=1$
$\Rightarrow \tan A+\tan B=1-\tan A\tan B$
$\Rightarrow \frac{x-1}{x-2}+\frac{x+1}{x+2}=1-\frac{x-1}{x-2}\times \frac{x+1}{x+2}$
$\Rightarrow (x-1)(x+2)+(x+1)(x-2)=(x-2)(x+2)-(x-1)(x+1)$
$\Rightarrow x^2+x-2+x^2-x-2=x^2-4-x^2+1$
$\Rightarrow 2x^2=1$
$\Rightarrow x=\pm \frac{1}{\sqrt{2}}$