You need to remember that the tangent function is rational, hence `tan x = sin x/cos x` .

You should come up with the substitution `arccos (-3/4) = pi - arccos (3/4) = y` such that:

`tan y = sin y/cos y`

`tan y = (sin(pi - arccos 3/4))/(cos(pi - arccos 3/4))`

You need to use the following formulas such that:

`sin (x - y) = sin x*cos y - sin y*cos x`

`cos (x - y) = cos x*cos y + sin x*sin y`

`tan y = (sin pi*cos(arccos 3/4) - sin(arccos 3/4)*cos pi)/(cos pi*cos (arccos 3/4) + sin pi*sin (arccos 3/4))`

You need to substitute 0 for `sin pi` and -1 for `cos pi` such that:

`tan y = (sin(arccos 3/4))/(-3/4)`

You need to remember that `sin(arccos x) = sqrt(1-x^2)` such that:

`tan y = (sqrt(1 - 9/16))/(-3/4)`

`tan y = (sqrt(7/16))/(-3/4) =gt tan y = -(sqrt7)/3`

**Hence, evaluating the tangent of `arccos(-3/4)` yields tan`(arccos(-3/4)) = -(sqrt7)/3` .**