How do you show arctan (x) ± arctan (y) = arctan [(x ± y) / (1 ± xy)] arctan(x)±arctan(y)=arctan[x±y1±xy]?

1 Answer
mason m
May 25, 2016

First, we should state the tangent addition formula:

tan(alpha+-beta)=(tan(alpha)+-tan(beta))/(1∓tan(alpha)tan(beta))tan(α±β)=tan(α)±tan(β)1∓tan(α)tan(β)

Rearrange by taking the arctangent of both sides:

alpha+-beta=arctan((tan(alpha)+-tan(beta))/(1∓tan(alpha)tan(beta)))α±β=arctan(tan(α)±tan(β)1∓tan(α)tan(β))

Now, let:

  • alpha=arctan(x)" "=>" "x=tan(alpha)α=arctan(x) ⇒ x=tan(α)
  • beta=arctan(y)" "=>" "y=tan(beta)β=arctan(y) ⇒ y=tan(β)

Make the substitutions into the tangent formula:

arctan(x)+-arctan(y)=arctan((x+-y)/(1∓xy))arctan(x)±arctan(y)=arctan(x±y1∓xy)

So, your identity is a little bit off since the minus-plus sign (∓∓) is needed in the denominator instead of the plus-minus (pm±) sign. The minus-plus sign shows that the identity can be split as follows:

{(arctan(x)+arctan(y)=arctan((x+y)/(1-xy))),(arctan(x)-arctan(y)=arctan((x-y)/(1+xy))):}

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