Question

How do you prove `tan(x + (pi/2)) = -cotx`?

Answer 1

We can not simply use the tangent of a sum formula, because `tan (pi/2)` does not exist. See other answer for using the sum formula.

So use
`tan theta = sin theta / cos theta` and the sum formulas for sine and cosine.

`tan(x + pi/2) = (sin(x + pi/2))/(cos(x + pi/2))`
`= (sinx cos(pi/2)+cosx sin(pi/2))/(cosx cos(pi/2)-sinx sin(pi/2)) = (0+cosx)/(0-sinx ) = - cosx/sinx = -cotx`

Answer 2

If we really want to use the sum formula for tangent, then we can. See below.

Explanation:

We cannot simply apply the sum formula as `tan(x+pi/2) = (tanx+tan(pi/2))/(1-tanxtan(pi/2)` because `tan(pi/2)` does not exist.

But we can rewrite `x+pi/2 = (x+pi/4)+pi/4` and apply the sun formula twice.

For any `a` in the domain of the tangent function, we have:

`tan(a+pi/4) = (tana+tan(pi/4))/(1-tanatan(pi/4))`

`= (tana+1)/(1-tana)`

Therefore,
`tan(x+pi/2) = tan((x+pi/4)+pi/4)`

`= (tan(x+pi/4)+1)/(1-tan(x+pi/4))`

`= ([(tanx+1)/(1-tanx)]+1)/(1-[(tanx+1)/(1-tanx)])`

`= ([(tanx+1)/(1-tanx)]+1)/(1-[(tanx+1)/(1-tanx)])*(1-tanx)/(1-tanx)`

`= (tanx+1+1-tanx)/(1-tanx-(tanx+1))`

`= 2/(-2tanx)`

`= -cotx`