If 1x(x2+a2)=Ax+Bx+Cx2+a2 then tan−1(AB)=

1=A(x2+a2)+(Bx+c)xPut x=0⇒A=1a2Coefficient of x2⇒A+B=0⇒B=−1a2Coefficient of x⇒c=0 tan−1(AB)=tan−1(−1)=−π4

If 1x(x2+a2)=Ax+Bx+Cx2+a2 then tan−1(AB)=

1=A(x2+a2)+(Bx+c)xPut x=0⇒A=1a2Coefficient of x2⇒A+B=0⇒B=−1a2Coefficient of x⇒c=0 tan−1(AB)=tan−1(−1)=−π4

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If $\frac{1}{x (x^{2} + a^{2})} = \frac{A}{x} + \frac{B x + C}{x^{2} + a^{2}}$ 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

$1 = A (x^{2} + a^{2}) + (B x + c) x$
Put $x = 0 \Rightarrow A = \frac{1}{a^{2}}$
Coefficient of $x^{2} \Rightarrow A + B = 0 \Rightarrow B = \frac{- 1}{a^{2}}$
Coefficient of $x \Rightarrow c = 0$ $\tan^{- 1} (\frac{A}{B}) = \tan^{- 1} (- 1) = - \frac{π}{4}$