the question and answer are there itself. what is required?

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the question and answer are there itself. what is required?

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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)

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.

======================================================

Let L H S = R H S = Ф

As L H S = Ф, tan Ф = (1-x) / (1- x)