If frac{3x+1}{(x-1)(x+3)}=frac{A}{x-1}+frac{B | vedclass

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(C) Given,$\frac{3x+1}{(x-1)(x+3)}=\frac{A}{x-1}+\frac{B}{x+3}$Multiplying both sides by $(x-1)(x+3)$,we get:$3x+1 = A(x+3) + B(x-1)$Expanding the rig

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$\frac{\pi}{6}$

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(C) Given,$\frac{3x+1}{(x-1)(x+3)}=\frac{A}{x-1}+\frac{B}{x+3}$Multiplying both sides by $(x-1)(x+3)$,we get:$3x+1 = A(x+3) + B(x-1)$Expanding the right side:$3x+1 = (A+B)x + (3A-B)$Comparing the coefficients of $x$ and the constant terms:$A+B = 3$ $(i)$$3A-B = 1$ (ii)Adding equations $(i)$ and (ii):$(A+B) + (3A-B) = 3+1$$4A = 4 \Rightarrow A = 1$Substituting $A=1$ into equation $(i)$:$1+B = 3 \Rightarrow B = 2$Now,calculate $\sin^{-1} \frac{A}{B}$:$\sin^{-1} \frac{1}{2} = \frac{\pi}{6}$

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

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