The number of solutions of tan^{-1}left(x+fra | vedclass

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(B) Given equation: $	an^{-1}\left(x+\frac{2}{x}ight) - 	an^{-1}\left(x-\frac{2}{x}ight) = 	an^{-1}\left(\frac{4}{x}ight)$.Using the identity

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$2$

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(B) Given equation: $	an^{-1}\left(x+\frac{2}{x}ight) - 	an^{-1}\left(x-\frac{2}{x}ight) = 	an^{-1}\left(\frac{4}{x}ight)$.Using the identity $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$,let $A = x+\frac{2}{x}$ and $B = x-\frac{2}{x}$.Then $A-B = (x+\frac{2}{x}) - (x-\frac{2}{x}) = \frac{4}{x}$.And $1+AB = 1 + (x+\frac{2}{x})(x-\frac{2}{x}) = 1 + (x^2 - \frac{4}{x^2}) = x^2 - \frac{4}{x^2} + 1$.So,$	an^{-1}\left(\frac{4/x}{1 + x^2 - 4/x^2}ight) = 	an^{-1}\left(\frac{4}{x}ight)$.This implies $\frac{4/x}{1 + x^2 - 4/x^2} = \frac{4}{x}$.Assuming $x 
eq 0$,we divide by $4/x$ to get $1 + x^2 - 4/x^2 = 1$,which simplifies to $x^2 - 4/x^2 = 0$.Thus $x^4 = 4$,so $x^2 = 2$,which gives $x = \pm \sqrt{2}$.Both values satisfy the original equation. Therefore,there are $2$ solutions.

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 number of solutions of $\tan^{-1}\left(x+\frac{2}{x}\right) - \tan^{-1}\left(\frac{4}{x}\right) - \tan^{-1}\left(x-\frac{2}{x}\right) = 0$ is:

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