Question

If A = `1/pi [(sin^-1(xpi), tan^-1(x/pi)),(sin^-1(x/pi), cot^-1(pix))]`, B = `1/pi [(-cos^-1(x/pi), tan^-1 (x/pi)),(sin^-1(x/pi),-tan^-1(pix))]`, then A – B is equal to ______.

पर्याय

• I

• O

• 2I

• `1/2"I"`

Answer 1

If A = `1/pi [(sin^-1(xpi), tan^-1(x/pi)),(sin^-1(x/pi), cot^-1(pix))]`, B = `1/pi [(-cos^-1(x/pi), tan^-1 (x/pi)),(sin^-1(x/pi),-tan^-1(pix))]`, then A – B is equal to `1/2"I"`.

Explanation:

Given that: A = `1/pi [(sin^-1(xpi), tan^-1(x/pi)),(sin^-1(x/pi), cot^-1(pix))]`

And B = `1/pi [(-cos^-1(x/pi), tan^-1 (x/pi)),(sin^-1(x/pi),-tan^-1(pix))]`

A – B = `1/pi [(sin^-1(xpi), tan^-1(x/pi)),(sin^-1(x/pi), cot^-1(pix))] - 1/pi[(-cos^-1(xpi), tan^-1(x/pi)),(sin^-1(x/pi), -tan^-1(pix))]`

= `1/pi [(sin^-1(xpi) + cos^-1(xpi), tan^-1(x/pi) - tan^-1(x/pi)),(sin^-1(x/pi) -sin^-1(x/pi), cot^-1(pix) + tan^-1(pix))]`

= `1/pi[(pi/2, 0),(0, pi/2)]`  ......`[(because sin^-1x + cos^-1x = pi/2),(tan^-1x + cot^-1x = pi/2)]`

= `1/pi xx pi/2 [(1, 0),(0, 1)]`

= `1/2"I"`