If `A=1/pi[sin^(-1)(pix)tan^(-1)(x/pi)sin^(-1)(x/pi)cot^(-1)(pix)]` and `B=1/pi[-cot^(-1)(pix)tan^(-1)(x/pi)sin^(-1)(x/pi)-tan^(-1)(pix)]` , then `A-B` is equal to `I` (b) 0 (c) `2I` (d) `1/2I`

Options

A. I
B. 0
C. 2I
D. `(1)/(2)I`

Explanation

We have, `A=[{:((1)/(2)sin^(-1)xpi,(1)/(pi)tan^(-1)(x)/(pi)),((1)/(pi)sin ^(-1)(x)/(pi),(1)/(pi)cot^(-1)pix):}]`
and `B=[{:((-1)/(pi)cos^(-1)xpi,(1)/(pi)tan^(-1)(x)/(pi)),((1)/(pi)sin^(-1)(x)/( pi),(-1)/(pi)tan^(-1)pix):}]`
`therefore A-B=[{:((1)/(pi)(sin^(-1)xpi+cos^(-1)xpi),(1)/(pi)(tan^(-1)(x)/(pi)-tan^(-1)(x)/(pi)),((1)/(pi)(sin^(-1)(x)/(pi)-sin^(-1)(x)/(pi)),(1)/( pi)cot^(-1)pix+tan^(-1)pix):}]`
`=[{:((1)/(pi),(pi)/(2),0),(0,(1)/(pi),(pi)/(2)):}] [because sin^(-1)x+cos^(-1)x=(pi)/(2) "and" tan^(-1)x+cot^(-1)x=(pi)/(2)]`
`=(1)/(2)[{:(1,0),(0,1):}]`
`=(1)/(2)I`