Find tan(arccos[x+1]) w/o trigonometric functions or inve

[king&i]

Find an expression for tan(arccos[x+1]) that does not involve trigonometric functions or their inverses

If anyone can help, I would really appreciate it.

[stapel_eliz]

Draw a right triangle. The particular shape doesn't matter.

right triangle:
                    /|
                 /   |
              /      |
           /         |
        /            |
     /               |
  /                  |
----------------------

By definition, the arccosine returns an angle value. So arccos(x + 1) = @ for some angle @. (I'm using "@" to stand for "theta".) You need to find the tangent of that angle. So let's label the triangle with the angle:

right triangle:
                    /|
                 /   |
              /      |
           /         |
        /            |
     /               |
  /@                 |
----------------------

We also know, by definition, that "arccos(x + 1) = @" means that "cos(@) = x + 1". In this case, it will be more useful to express this as "(x + 1)/1", because that gives us our "adjacent" (namely, x + 1) and our "hypotenuse" (namely, 1). So now we have:

right triangle:
                    /|
                 /   |
         1    /      |
           /         |
        /            |
     /               |
  /@                 |
----------------------
         x + 1

Use the Pythagorean Theorem to find the value for "opposite", and then read off the value of the tangent.