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If $\frac{3 x+1}{(x-1)(x+3)}=\frac{A}{x-1}+\frac{B}{x+3}$, then $\sin ^{-1} \frac{A}{B}$ is equal to
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The correct answer is:
$\frac{\pi}{6}$
Given, $\frac{3 x+1}{(x-1)(x+3)}=\frac{A}{x-1}+\frac{B}{x+3}$
$$
\begin{aligned}
&\Rightarrow \quad 3 \mathrm{x}+1=\mathrm{A}(\mathrm{x}+3)+\mathrm{B}(\mathrm{x}-1) \\
&\Rightarrow \quad 3 \mathrm{x}+1=(\mathrm{A}+\mathrm{B}) \mathrm{x}+(3 \mathrm{~A}-\mathrm{B})
\end{aligned}
$$
On equating the coefficient of $x$ from both sides, we get
and $\quad \begin{array}{rr}\mathrm{A}+\mathrm{B} & =3 \\ 3 \mathrm{~A}-\mathrm{B} & =1\end{array}$
On adding Eqs. (i) and (ii), we get
$$
4 \mathrm{~A}=4 \Rightarrow \mathrm{A}=1
$$
$\therefore$ From Eq. (i), we get
$$
\begin{aligned}
B=3-A &=3-1=2 \\
\therefore \sin ^{-1} \frac{A}{B}=\sin ^{-1}\left(\frac{1}{2}\right) &=\frac{\pi}{6}
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow \quad 3 \mathrm{x}+1=\mathrm{A}(\mathrm{x}+3)+\mathrm{B}(\mathrm{x}-1) \\
&\Rightarrow \quad 3 \mathrm{x}+1=(\mathrm{A}+\mathrm{B}) \mathrm{x}+(3 \mathrm{~A}-\mathrm{B})
\end{aligned}
$$
On equating the coefficient of $x$ from both sides, we get
and $\quad \begin{array}{rr}\mathrm{A}+\mathrm{B} & =3 \\ 3 \mathrm{~A}-\mathrm{B} & =1\end{array}$
On adding Eqs. (i) and (ii), we get
$$
4 \mathrm{~A}=4 \Rightarrow \mathrm{A}=1
$$
$\therefore$ From Eq. (i), we get
$$
\begin{aligned}
B=3-A &=3-1=2 \\
\therefore \sin ^{-1} \frac{A}{B}=\sin ^{-1}\left(\frac{1}{2}\right) &=\frac{\pi}{6}
\end{aligned}
$$
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