Question

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

Answer 1

First, we should state the tangent addition formula:

`tan(alpha+-beta)=(tan(alpha)+-tan(beta))/(1∓tan(alpha)tan(beta))`

Rearrange by taking the arctangent of both sides:

`alpha+-beta=arctan((tan(alpha)+-tan(beta))/(1∓tan(alpha)tan(beta)))`

Now, let:

• `alpha=arctan(x)" "=>" "x=tan(alpha)`
• `beta=arctan(y)" "=>" "y=tan(beta)`

Make the substitutions into the tangent formula:

`arctan(x)+-arctan(y)=arctan((x+-y)/(1∓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))):}`