If sin^-1 ( 2a1+a^2 )+ cos^-1 ( 1-a^21+a^2 )= tan^-1 ( 2x1-x^2 ) where a ,x ∈ (0 ,1 ) then the value of x is

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Correct answer: C. (c) : We have, sin^-1 ⁡ ( 2a1+a^2 )+ cos^-1 ⁡ ( 1-a^21+a^2 ) = tan^-1 ⁡ ( 2x1-x^2 ) ,a ,x ∈ (0 ,1 ) ⇒ 2 tan^-1 ⁡a+2 tan^-1 ⁡a=2 tan^-1 ⁡x ( ∵ sin^-1 ⁡ ( 2x1+x^2 ) ) =2 tan^-1 ⁡x ,-1 x 1 cos^-1 ⁡ ( 1-x^21+x^2 ) =2 tan^-1 ⁡x ,0 x< tan^-1 ⁡ ( 2x1-x^2 )(=2 tan^-1 ⁡x ,-1<x<1) ⇒ 4 tan^-1 ⁡a=2 tan^-1 ⁡x ⇒2 tan^-1 ⁡a= tan^-1 ⁡x ⇒ tan^-1 ⁡ ( 2a1-a^2 )= tan^-1 ⁡x ⇒x= 2a1-a^2

$If sin^{- 1} (\frac{2 a}{1 + a ^{2}}) + cos^{- 1} (\frac{1 - a ^{2}}{1 + a ^{2}}) = tan^{- 1} (\frac{2 x}{1 - x ^{2}}) where a, x \in (0, 1) then the value of x is$

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