# Trigonometric identity for inverse tan

1. Apr 28, 2010

### Kat007

Hello,

Could you please clarify if this is correct:

If tan^(-1)(x) = Pi/2 - tan^(-1)(1/x)

Then if we have (ax) as the angle where a is a constant, do we get:
tan^(-1)(ax) = Pi/2 - tan^(-1)(a/x)
or does the constant go on the bottom with the x? i.e. or:
tan^(-1)(ax) = Pi/2 - tan^(-1)(1/(ax))

Thank you very much,
Kat

2. Apr 28, 2010

### Staff: Mentor

I believe that you can answer this yourself by drawing a right triangle (use a 30-60-30 triangle), and writing the tangent in terms of the ratio of the opposite and adjacent sides. Then double the height of the opposite side....

3. Apr 28, 2010

### Gigasoft

Do you think the identity should change if you replace the letter x with the letter y? Is the identity invalid for some value of y? What if y = ax? If you agree that it is still valid, what do you get when you put ax in the place of y?

4. Apr 28, 2010

### Kat007

OK so it seems to work out as the
tan^(-1)(ax) = Pi/2 - tan^(-1)(1/(ax))
Could you please confirm that this is correct?

5. Apr 28, 2010

### Staff: Mentor

Which method did you use to conclude that? Giga and I have a bet going...

6. Apr 28, 2010

### Kat007

lol how kind of you.. Jokers heheee
The triangle one.
They both makes sence. Thanks,
Kat

7. Apr 28, 2010

### Staff: Mentor

Oh fooey. If they both make sense, then neither of us wins the bet. Oh well

8. Apr 28, 2010

### Mentallic

Nah, Giga's make more sense AND no pen and paper required :tongue: