Question

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Answer 1

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Answer 2

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Let, tan^-1 (x) = θ         

Therefore, tan θ = x

We know that,

tan 2θ = 2tanθ/(1−tan^2θ)

tan 2θ = 2x/(1−x^2)

2θ = tan−1−1(2x1−x22x1−x2)

2 tan^-1(x) = tan^-1 {2x /(1−x^2)} ……. 1

Again,

sin 2θ = 2tanθ/1+tan^2(θ)

sin 2θ = 2x1+x22x1+x2

2θ = sin^-1 {2x / (1+x^2)}

2 tan^-1= sin^-1{2x/(1+x^2)} …………………….. 2

Now,  cos 2θ = 1−tan^2(θ)/1+tan^2(θ)

 cos 2θ = ( 1−x^2)/ (1+x^2)

2θ = cos^-1 (1−x^2/1+x^2)

2 tan^-1(x) = cos{ (1−x^2) /(1+x^2)}……………………..3

Therefore, from (i), (ii) and (iii)
we get           

2Tan^-1(x)= Sin^-1{2x/(1+x^2)} =Cos^-1 (1-x^2)/(1+x^2) = tan^-1( 2x/ 1-x^2)

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