# Real x, y verify relation arctg x+arctg y=pi/2. How show x*y=1?

*print*Print*list*Cite

### 1 Answer

You should move one member to the right side, such that:

`arctan x = pi/2 - arctan y`

You need to take the tangent function both sides, such that:

`tan(arctan x) = tan(pi/2 - arctan y)`

You need to use the following trigonometric identity `cot alpha = ` `tan(pi/2 - alpha)` , such that:

`tan(pi/2 - arctan y) = cot (arctan y)`

You need to also use the following trigonometric identity, such that:

`cot alpha = 1/(tan alpha)`

Reasoning by analogy, yields:

`cot (arctan y) = 1/(tan(arctan y))`

Since ` tan(arctan alpha) = alpha` , you need to replace `x` for `tan(arctan x)` and y for `(tan(arctan y))` :

`x = 1/y`

You need to test if `x*y = 1` , under the given conditions, hence, replacing `1/y` for `x ` yields:

`1/y*y = y/y = 1`

**Hence, checking if the relation `x*y = 1` holds, under the given conditions, yields that the relation `x*y = 1` is valid.**