The possible values of x, which satisfy the t | vedclass

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(A) Given the equation: $	an ^{-1}\left(\frac{x-1}{x-2}ight)+	an ^{-1}\left(\frac{x+1}{x+2}ight)=\frac{\pi}{4}$ Using the formula $	an ^{-1} A

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

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(A) Given the equation: $	an ^{-1}\left(\frac{x-1}{x-2}ight)+	an ^{-1}\left(\frac{x+1}{x+2}ight)=\frac{\pi}{4}$ Using the formula $	an ^{-1} A + 	an ^{-1} B = 	an ^{-1} \left( \frac{A+B}{1-AB} ight)$, we get: $	an ^{-1}\left[\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\left(\frac{x-1}{x-2}ight) \left(\frac{x+1}{x+2}ight)}ight]=\frac{\pi}{4}$ Taking $	an$ on both sides: $\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{(x-1)(x+1)}{(x-2)(x+2)}} = 	an \frac{\pi}{4} = 1$ Simplifying the numerator and denominator: $\frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)-(x-1)(x+1)} = 1$ $\frac{(x^2+x-2)+(x^2-x-2)}{(x^2-4)-(x^2-1)} = 1$ $\frac{2x^2-4}{-3} = 1$ $2x^2-4 = -3$ $2x^2 = 1$ $x^2 = \frac{1}{2}$ $x = \pm \frac{1}{\sqrt{2}}$

List-$I$	List-$II$
$A$. $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\sin ^{-1} \frac{1}{3}$	$I$. $k \pi \pm(-1)^k \frac{\pi}{6}, k \in Z$
$B$. $\sin ^{-1}\left(\frac{(-1)^n}{2}\right), n \in Z$	$II$. $k \pi \pm 1, k \in Z$
$C$. $\tan ^{-1}\left(\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right)$	$III$. $\frac{3}{2}$
$D$. $\sin ^{-1}\|\sin x\|=\sqrt{\sin ^{-1}\|\sin x\|} \Rightarrow x \in$	$IV$. $\frac{3 \pi}{8}$
	$V$. $\frac{\pi}{2}$

The possible values of $x$, which satisfy the trigonometric equation $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$ are

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