Question

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

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

Answer 1

Answer: (A)

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