Question

If $\frac{3x+1}{(x-1)(x+3)}=\frac{A}{x-1}+\frac{B}{x+3}$, then ${\sin }^{-1}\frac{A}{B}=$

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is C $\frac{\pi }{6}$

We have, $\frac{3x+1}{(x-1)(x+3)}=\frac{A}{x-1}+\frac{B}{x+3}$,

$\Rightarrow 3x+1=A(x+3)+B(x-1)$

Substitute $x=1$ both sides, $\Rightarrow 3\left(1\right)+1=A(1+3)+0\Rightarrow A=1$

Substitute $x=-3$ both sides, $\Rightarrow 3(-3)+1=0+B(-3-1)$

$\Rightarrow -8=-4B\Rightarrow B=2$

Hence ${\sin }^{-1}\frac{A}{B}={\sin }^{-1}\frac{1}{2}=\frac{\pi }{6}$