# Inverse tangent proof

1. Oct 18, 2006

### m00c0w

Show that $$\arctan{x} + \arctan {y} = \arctan { \frac{x+y}{1-xy} }$$ when $$x = \frac{1}{2}\ and \y = \frac{1}{3}$$ but not when $$x = 2\ and \y = 3$$

I've tried taking the tangent of both sides but I don't know what to do then when I've got $$\tan ( \arctan{x} + \arctan{y} ) = \frac{x+y}{1-xy}$$

Any help would be greatly appreciated. Thanks!

Last edited: Oct 18, 2006
2. Oct 18, 2006

### courtrigrad

Use the fact that $$\tan(u+v) = \frac{\tan u + \tan v}{1-\tan u \tan v}$$

3. Oct 19, 2006

### HallsofIvy

Staff Emeritus
I don't see what the difficulty is. You are aske to show that this equation is satisfied when $x= \frac{1}{2}$ and $y= \frac{1}{3}$ but not when x= 2 and y= 3.

Okay, plug those values in and evaluate. Courtrigrax's method would appear to be a general method of proving that it is an identity: true for all x and y which contradicts the problem!

4. Oct 19, 2006

### HallsofIvy

Staff Emeritus
I'm not clear what the problem is. You are asked to show that this equation is satisfied when $x= \frac{1}{2}$ and $y= \frac{1}{3}$ but not when x= 2 and y= 3.

Okay, plug those values in and evaluate.

Courtrigrax's method would appear to be a general method of proving that it is an identity: true for all x and y, which contradicts the statement of the problem!

5. Oct 19, 2006

### m00c0w

Ok, so I used the identity $$\tan(u+v) = \frac{\tan u + \tan v}{1-\tan u \tan v}$$ but all it does is bring me to the equation $$\frac{x+y}{1-xy} = \frac{x+y}{1-xy}$$. I had tried substituting the values in before and found that the statement held true for both sets of x and y, but I assumed I must have done something wrong. I told my teacher and she said I was wrong

So have I messed up? Or does the equation hold true for x = 2 and y = 3 thus rendering the proof impossible?

6. Oct 19, 2006

### courtrigrad

The equation does not hold for x = 2 and y = 3. Just plug in the values. Also, can you see that it will work for $$x > -1$$ , $$y < 1$$? Why is this?

It is because the left hand side is not defined if $$xy = 1$$

Last edited: Oct 19, 2006
7. Oct 20, 2006

### m00c0w

I plugged in the values and got the following...

$$\arctan {2}\ + \arctan {3} = \arctan {-1}$$
$$2.35619... = -0.78539...$$

However another solution to $$\arctan {-1}$$ is $$-0.78539... + pi = 2.35619...$$ which makes the statement hold true Am I not allowed to take anything other than the principal value? If so, why not?

Also, you said that $$x > -1$$, $$y < 1$$. Are you saying that these are the only values of x and y that the equation will hold for? That doesn't seem to make sense, as if, for example, I use x = -5 and y = 10 the equation will still hold true.

Last edited: Oct 20, 2006
8. Oct 20, 2006

### HallsofIvy

Staff Emeritus
tangent(x) has a discontinuity at $\pi/2$ or approximately 1.57.

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