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If $\sin ^{-1}\left(\frac{2 p}{1+p^{2}}\right)-\cos ^{-1}\left(\frac{1-q^{2}}{1+q^{2}}\right)$ $=\tan ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$, then the value of $x$ is equal to
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Verified Answer
The correct answer is:
$\frac{p-q}{1+p q}$
We have,
$$
\begin{array}{ll}
\sin ^{-1}\left(\frac{2 p}{1+p^{2}}\right)-\cos ^{-1}\left(\frac{1-q^{2}}{1+q^{2}}\right) \\
& =\tan ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \\
\Rightarrow & 2 \tan ^{-1} p-2 \tan ^{-1} q=2 \tan ^{-1} x \\
\Rightarrow & 2\left(\tan ^{-1} p-\tan ^{-1} q\right)=2 \tan ^{-1} x \\
\Rightarrow \quad & \tan ^{-1} p-\tan ^{-1} q=\tan ^{-1} x \\
\Rightarrow & \tan ^{-1}\left(\frac{p-q}{1+p q}\right)=\tan ^{-1} x \\
\Rightarrow & x=\frac{p-q}{1+p q}
\end{array}
$$
$$
\begin{array}{ll}
\sin ^{-1}\left(\frac{2 p}{1+p^{2}}\right)-\cos ^{-1}\left(\frac{1-q^{2}}{1+q^{2}}\right) \\
& =\tan ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \\
\Rightarrow & 2 \tan ^{-1} p-2 \tan ^{-1} q=2 \tan ^{-1} x \\
\Rightarrow & 2\left(\tan ^{-1} p-\tan ^{-1} q\right)=2 \tan ^{-1} x \\
\Rightarrow \quad & \tan ^{-1} p-\tan ^{-1} q=\tan ^{-1} x \\
\Rightarrow & \tan ^{-1}\left(\frac{p-q}{1+p q}\right)=\tan ^{-1} x \\
\Rightarrow & x=\frac{p-q}{1+p q}
\end{array}
$$
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