Question

If $\frac {(x+1)^2}{x^3+x}= \frac {A}{x}+\frac {Bx+C}{x^2 +1} $ then $\cos ec^{-1}$ $(\frac {1}{A}) + \cot^{-1} \frac {1}{B}+\sec^{-1}C$ is equal to

• A. $\frac {5 \pi} {6}$
• B. $0$
• C. $\frac {\pi} {6}$
• D. $\frac { \pi} {2}$

Answer 1

Answer: (A)

$\frac{(x+1)^{2}}{x^{3}+x} =\frac{x^{2}+1+2 x}{x\left(x^{2}+1\right)}=\frac{x^{2}+1}{x\left(x^{2}+1\right)}+\frac{2 x}{x\left(x^{2}+1\right)} $
$=\frac{1}{x}+\frac{2}{x^{2}+1}$
On comparing, $A=1, B=0$ and $C=2$
Now, $cosec^{-1}\left(\frac{1}{A}\right)+\cot ^{-1}\left(\frac{1}{B}\right)+\sec ^{-1} C$
$=\cos ec^{-1}(1)+\cot ^{-1}\left(\frac{1}{0}\right)+\sec ^{-1} 2 $
$=\frac{\pi}{2}+0+\frac{\pi}{3}=\frac{5 \pi}{6}$