If ⁻¹ ( 2 a 1+a^2 )+ ⁻¹ ( 1-a^2 1+a^2 )= ⁻¹ ( 2 x 1-x^2 ) where a, x (0,1) , then the value of x is

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Correct answer: C. 2 a 1-a^2

Here, ⁻¹ ( 2 a 1+a^2 )+ ⁻¹ ( 1-a^2 1+a^2 )= ⁻¹ ( 2 x 1-x^2 ) We know that 2 ⁻¹ x= ⁻¹ ( 2 x 1+x^2 ) and 2 ⁻¹ x= ⁻¹ ( 1-x^2 1+x^2 ) 2 ⁻¹ a+2 ⁻¹ a= ⁻¹ ( 2 x 1-x^2 )2 [ ⁻¹ ( a+a 1-a a ) ]=2 ⁻¹ x 2 a 1-a^2 =x

If ⁻¹ ( 2 a 1+a^2 )+ ⁻¹ ( 1-a^2 1+a^2 )= ⁻¹ ( 2 x 1-x^2 ) where a, x (0,1) , then the value of x is

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