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$2 S i n^{- 1} x = S i n^{- 1} (2 x \sqrt{1 - x^{2}})$ holds good for

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a

x∈[0,1]

b

x∈[−1,1]

c

x∈−12,12

d

x∈[−1,0]

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detailed solution

Correct option is C

Let sin-1x=θ, then x∈-1,1 and θ∈-π2,π2. ⇒sinθ=x ⇒cosθ=1-sin2θ=1-x2 ∵θ∈-π2,π2⇒cosθ is positive Now sin2θ=2sinθcosθ=2x1-x2 ⇒2ϑ=sin-12x1-x2 if 2ϑ∈-π2,π2 by definition of sin-1 ⇒2sin-1x=sin-12x1-x2 when ϑ∈-π4,π4 Now -π4<θ<π4⇒sin-π4

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