Question

If (x+1)^2/x^3+x = A/x + Bx+C/x^2+1, then sin^-1 A + tan^-1 B +sec^-1 C =

1) pi/2
2) pi/6
3) 0
4) 5pi/6

Answer 1

Answer:

4) 5π/6

Step-by-step explanation:

We have,

$\frac{ {(x + 1)}^{2} }{ {x}^{3} + x} = \frac{a}{x} + \frac{bx + c}{ {x}^{2} + 1 }$

$= > \frac{ {x}^{2} + 2x + 1}{x( {x}^{2} + 1) } = \frac{a( {x}^{2} + 1) + x(bx + c)}{x( {x}^{2} + 1)}$

$= > {x}^{2} + 2x + 1 = (a + b) {x}^{2} + cx + a$

On comparing both sides,

a = 1, c = 2, b = 0

so,

$\sin^{ - 1} (a) + \tan ^{ - 1} (b) + \sec^{ - 1} (c) = \sin^{ - 1} (1) + \tan ^{ - 1} (0) + \sec^{ - 1} (2)$

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

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