Identity Proofs of Inverse Trig Functions

[MHrtz]

Homework Statement

Prove the Identity (show how the derivatives are the same):

arcsin ((x - 1)/(x + 1)) = 2arctan (sqr(x) - pi/2)

Homework Equations

d/dx (arcsin x) = 1/ sqr(1 - x²)

d/dx (arctan x) = 1/ (1 + x²)

All my attempts have been messy and it may be because I didn't take the derivatives properly.
I attached what I got for the derivatives for both. If it's not the right derivative than what is? If it is the right derivative, where do I go from here?

 

[micromass]

Hi MHrtz!

Are you sure that that identity even holds? It doesn't seem to hold in the case x=1...

 

[MHrtz]

Are you suggesting that the derivatives of these functions are incorrect? I got them from a textbook so I assumed they were correct.

 

[micromass]

Try to substitute x=1 in

arcsin ((x - 1)/(x + 1)) = 2arctan (sqr(x) - pi/2)

 

[MHrtz]

I get different answers for both sides but for an identity this doesn't matter. What about my derivation? Is it correct?

 

[micromass]

I get different answers for both sides but for an identity this doesn't matter.

An identity means that something holds for all x, no?

What about my derivation? Is it correct?

Yes, I think that is correct.

 

[MHrtz]

I just realized I wrote the problem down wrong. the pi/2 is outside of the arctan. I need to redo my derivative.

 

[MHrtz]

So here is the corrected derivative. How can I use it to prove the identity?

 

[micromass]

Seems like the correct derivative. Now you need to rewrite it until it becomes obvious that it's equal. Note that your equation is certainly equivalent with

$$\frac{\sqrt{1-\left(\frac{x-1}{x+1}\right)^2}(x+1)^2}{2}=\sqrt{x}(x+1)$$

Now, maybe try to square both sides?

 

[MHrtz]

I tried simplifying but then this happened.

 

[micromass]

The mistake is in your third step. That should read

$$\frac{\left(1-\left(\frac{x-1}{x+1}\right)^2\right)(x+1)^2}{4}=x$$

You forgot some brackets...

 

[MHrtz]

ok I simplified both sides to where x = x. Thank You, I can take it from here.