Question

If $\frac{(x+1{)}^2}{x^3+x}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$, then ${\sin }^{-1}A+{\tan }^{-1}B+{\sec }^{-1}C=$

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

Answer 1

Answer: (D)

$\frac{5\pi }{6}$

We have: $\frac{(x+1{)}^2}{x^3+x}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$

$\Rightarrow \frac{(x+1{)}^2}{x^3+x}=\frac{A(x^2+1)+x(Bx+C)}{x^3+x}$

$\Rightarrow x^2+2x+1=A{x}^2+A+B{x}^2+Cx$

$\Rightarrow A+B=1,C=2,A=1$

$\therefore B=0$

$\Rightarrow {\sin }^{-1}A+{\tan }^{-1}B+{\sec }^{-1}C={\sin }^{-1}1+{\tan }^{-1}0+{\sec }^{-1}2$

$=\frac{\pi }{2}+0+\frac{\pi }{3}$$=\frac{5\pi }{6}$