The number of solutions of operatorname{Tan}^ | vedclass

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(D) Given equation: $\operatorname{Tan}^{-1} 1 + \frac{1}{2} \operatorname{Cos}^{-1} x^2 - \operatorname{Tan}^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2

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infinitely many

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(D) Given equation: $\operatorname{Tan}^{-1} 1 + \frac{1}{2} \operatorname{Cos}^{-1} x^2 - \operatorname{Tan}^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}ight) = 0$. Let $x^2 = \cos 2	heta$,where $2	heta \in [0, \pi]$. Then $\sqrt{1+x^2} = \sqrt{1+\cos 2	heta} = \sqrt{2}\cos	heta$ and $\sqrt{1-x^2} = \sqrt{1-\cos 2	heta} = \sqrt{2}\sin	heta$. The third term becomes $\operatorname{Tan}^{-1}\left(\frac{\cos	heta+\sin	heta}{\cos	heta-\sin	heta}ight) = \operatorname{Tan}^{-1}\left(	an(\frac{\pi}{4}+	heta)ight) = \frac{\pi}{4} + 	heta$. Substituting back: $\frac{\pi}{4} + \frac{1}{2}(2	heta) - (\frac{\pi}{4} + 	heta) = 0$. This simplifies to $0 = 0$,which is an identity. However,the domain of the expression requires $1-x^2 \ge 0$,so $x^2 \le 1$,and the denominator $\sqrt{1+x^2}-\sqrt{1-x^2} 
eq 0$,so $x^2 
eq 1$. Also,$x^2 \ge 0$. Thus,$x^2 \in [0, 1)$. Since $x^2 = \cos 2	heta$,this corresponds to $2	heta \in (0, \pi/2]$,which means $x^2 \in (0, 1]$. Combining these,$x^2 \in (0, 1)$. Since there are infinitely many values of $x$ in the interval $(-1, 0) \cup (0, 1)$,the number of solutions is infinitely many.

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 $\operatorname{Tan}^{-1} 1 + \frac{1}{2} \operatorname{Cos}^{-1} x^2 - \operatorname{Tan}^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right) = 0$ is

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