# arctg(x+1) - arctg(x+1)=pi/12

Notice that the terms to the left side are equal, hence, performing the subtraction yields 0, hence, there is no x for `arctan(x+1) - arctan(x + 1) = pi/12` .

supposing that the terms to the left side are added, yields:

`arctan(x+1) + arctan(x + 1) = pi/12`

`2 arctan...

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Notice that the terms to the left side are equal, hence, performing the subtraction yields 0, hence, there is no x for `arctan(x+1) - arctan(x + 1) = pi/12` .

supposing that the terms to the left side are added, yields:

`arctan(x+1) + arctan(x + 1) = pi/12`

`2 arctan (x+1) = pi/12 => arctan(x+1) = (pi/12)/2 => arctan(x+1) = pi/24`

You need to evaluate tangent function of both sides, such that:

`tan(arctan(x+1)) = tan(pi/24)`

`x + 1 = tan(pi/24) => x = tan(pi/24) - 1= -0.99`

Hence, evaluating the original equation, yields 0!=pi/12 and if you solve the equation `arctan(x+1) + arctan(x + 1) = pi/12`  yields `x = -0.99` .

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