Proving Inverse Tangent Identity: x = 1/2, y = 1/3 vs. x = 2, y = 3

[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!

 

[courtrigrad]

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

 

[HallsofIvy]

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!

 

[HallsofIvy]

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!

 

[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?

 

[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

 

[m00c0w]

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

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.

 

[HallsofIvy]

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