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2tan^-1x = cos^-1(1-x^2/1+x^2)

tan-1x = y

x= tany

2tan-1x= cos-1 1-x2/1+x2

= cos-1 1-tan2y/1+tan2y

= cos-1(cos2y)

=2y

=2tan-1x

therefore, 2tan-1x = cos-1 1-x2/1+x2

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X^2 tan^-1x /1+x^2
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Write the set of values of x for which 2tan-1x = cos-1(1-x2/1+x2) holds
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x>=o
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Write the values of x for which 2tan-1x =cos-1(1-x2/1+x2)
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