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2014-08-20T06:13:41+05:30

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To solve    tan^-1 [ (1-x)/(1+x) ]  = 1/2 * tan^-1  x  ,    x > 0

This is easily solved using the following formula. That is done in the solution there, i guess. That is a simple solution.  You can in a little longer way without using that too.

                   tan (A - B) = (tan A - tan B) / (1 + tan A tan B)

      So  A - B  =  tan^-1 [ (tan A - tan B) / (1 + tan A tan B) ]

     comparing with LHS, we find  tan A = 1, so  A = π/4 ; tan B = x  So B = tan^-1 x

     So   A - B = π/4 - tan^-1 x  =  RHS = 1/2 tan^-1 x
     3/2 tan^-1 x  = π/4          tan^-1 x  = 2π/3*4 = π/6 
         x = tan π/6  = 1/√3
Easy way to get the solution.
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Let  L H S = R H S =  Ф
As L H S = Ф,     tan Ф  =  (1-x) / (1- x)

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