# Please show a short way to arccos( (1-x^2)/(1+x^2) ) = 2*arctan(x) for x out of [0,infinity)This can be tricky

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You need to form the function `f(x) =arccos((1-x^2)/(1+x^2)) -2*arctan(x) = 0` .

Since the function is constant, then the derivative cancels. Hence, you need to find the first derivative to verify is cancels.

`f'(x) = (-((1-x^2)/(1+x^2))')/sqrt(1 -((1-x^2)/(1+x^2))^2) - 2/(1 + x^2)`

You need to use quotient rule to find `((1-x^2)/(1+x^2))'` such that:

`(-((1-x^2)/(1+x^2))')/sqrt(1 -((1-x^2)/(1+x^2))^2) = ((-(1-x^2)'*(1+x^2) - (1-x^2)*(1+x^2)')/((1+x^2)^2))/(((sqrt((1+x^2)^2 - (1-x^2)^2))/(1+x^2))`

`(-((1-x^2)/(1+x^2))')/sqrt(1 -((1-x^2)/(1+x^2))^2) = (2x(1+x^2)+2x(1-x^2))/((1+x^2)sqrt(1+x^2-1+x^2)(1+x^2+1-x^2))`

`` `f'(x) = 4x/(2x(1+x^2)) - 2/(1 + x^2)`

Reducing by 2x yields:

`f'(x) = 2/(1+x^2) - 2/(1 + x^2) = 0`

**Since `f'(x) = 0` , hence the function `f(x) = arccos((1-x^2)/(1+x^2)) -2*arctan(x) = 0**

**arccos((1-x^2)/(1+x^2)) = 2*arctan(x)`**