If 1xx2+a2=Ax+Bx+Cx2+a2 then tan-1AB=

1x(x2+a2)=Ax+Bx+Cx2+a21=A(x2+a2)+(Bx+C)xx=0 ⇒ 1=A a2 ⇒ A=1a2x2 coeff ⇒ 0=A+B ⇒ B=-1a2 AB=-1 Tan-1AB=-π4

If 1xx2+a2=Ax+Bx+Cx2+a2 then tan-1AB=

1x(x2+a2)=Ax+Bx+Cx2+a21=A(x2+a2)+(Bx+C)xx=0 ⇒ 1=A a2 ⇒ A=1a2x2 coeff ⇒ 0=A+B ⇒ B=-1a2 AB=-1 Tan-1AB=-π4

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$If \frac{1}{x (x^{2} + a^{2})} = \frac{A}{x} + \frac{Bx + C}{x^{2} + a^{2}} then \tan^{- 1} [\frac{A}{B}] =$

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$\frac{3 π}{4}$

$\frac{π}{4}$

$- \frac{π}{4}$

$\frac{π}{3}$

answer is C.

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Detailed Solution

$\begin{array}{rcl} \frac{1}{x (x^{2} + a^{2})} & = & \frac{A}{x} + \frac{Bx + C}{x^{2} + a^{2}} \\ 1 & = & A (x^{2} + a^{2}) + (Bx + C) x \\ x & = & 0 \Rightarrow 1 = A a^{2} \Rightarrow A = \frac{1}{a^{2}} \\ x^{2} coeff & \Rightarrow & 0 = A + B \Rightarrow B = - \frac{1}{a^{2}} \\ \frac{A}{B} & = & - 1 \\ {Tan}^{- 1} (\frac{A}{B}) & = & - \frac{π}{4} \end{array}$