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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 ______.
Options
I
O
2I
`1/2"I"`
Solution
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"`
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