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If $\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}$, then $\sin ^{-1} A+\tan ^{-1} B+\sec ^{-1} C$ is equal to
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Verified Answer
The correct answer is:
$\frac{5 \pi}{6}$
Given,
$\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}...(i)$
$\begin{aligned} \Rightarrow \quad(x+1)^{2} &=A\left(x^{2}+1\right)+(B x+c)(x) \\ x^{2}+1+2 x &=A x^{2}+A+B x^{2}+C x \\ &=(A+B) x^{2}+C x+A \end{aligned}$
On comparing the coefficient of like powers on both sides, we get
$A+B=1...(ii)$
and $\quad C=2, \quad A=1$
From Eq. (ii), $B=0$
Then, $\sin ^{-1} A+\tan ^{-1} B+\sec ^{-1} C$
$=\sin ^{-1}(1)+\tan ^{-1}(0)+\sec ^{-1}(2)$
$=\sin ^{-1} \sin \frac{\pi}{2}+\tan ^{-1} \tan 0+\sec ^{-1} \sec \frac{\pi}{3}$
$=\frac{\pi}{2}+0+\frac{\pi}{3}=\frac{5 \pi}{6}$
$\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}...(i)$
$\begin{aligned} \Rightarrow \quad(x+1)^{2} &=A\left(x^{2}+1\right)+(B x+c)(x) \\ x^{2}+1+2 x &=A x^{2}+A+B x^{2}+C x \\ &=(A+B) x^{2}+C x+A \end{aligned}$
On comparing the coefficient of like powers on both sides, we get
$A+B=1...(ii)$
and $\quad C=2, \quad A=1$
From Eq. (ii), $B=0$
Then, $\sin ^{-1} A+\tan ^{-1} B+\sec ^{-1} C$
$=\sin ^{-1}(1)+\tan ^{-1}(0)+\sec ^{-1}(2)$
$=\sin ^{-1} \sin \frac{\pi}{2}+\tan ^{-1} \tan 0+\sec ^{-1} \sec \frac{\pi}{3}$
$=\frac{\pi}{2}+0+\frac{\pi}{3}=\frac{5 \pi}{6}$
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