Given that,
\[
\begin{aligned}
\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) &=\left\{\begin{array}...
Given that,
\[
\begin{aligned}
\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) &=\left\{\begin{array}{l}
2 \tan ^{-1} x,|x| \leq 1 \\
-\pi+2 \tan ^{-1} x, x1 \\
\pi+2 \tan ^{-1} x, x-1
\end{array}\right.\\
\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=\left\{\begin{array}{l}
2 \tan ^{-1} x,|x| \leq 1 \\
\pi-2 \tan ^{-1} x, x1 \text { and } \\
-\left(\pi+2 \tan ^{-1} x\right), x-1
\end{array}\right.
\end{aligned}
\]
P
W
\[
\sin ^{-1} x+\cos ^{-1} x=\pi / 2 \text { for }-1 \leq x \leq 1
\]
\( \sin ^{-1}\left(\frac{4 x}{x^{2}+4}\right)+2 \tan ^{-1}\left(-\frac{x}{2}\right) \) is independent of \( x \), then
(a) \( x \in[-3,4] \)
(b) \( x \in[-2,2] \)
(c) \( x \in[-1,1] \)
(d) \( x \in[1, \infty) \)
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