Converting one inverse trig function to another

[cr7einstein]

Homework Statement

Express $$ 2 arctan (\sqrt\frac{a-b}{a+b} tan (\theta/2))$$ in terms of inverse cosine

Homework Equations

I realize it amounts to find a smart substitution, but I can't find one.

The Attempt at a Solution

I tried $b/a=tan \theta$ , but I can't find any way to get rid of the other tangent term. I would be really nice if the solution is not a direct substitution of the enormous formula for the inverse trig functions, as it only gets messy. Thanks in advance!

 

[member 587159]

I don't have the time to look at it now, but here are some identities relating arcsin and arctan:https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions

 

[cr7einstein]

@Math_QED I mentioned that I don't want to just substitute into the formulae as they are very messy...No problem, you can look into it when you have the time.

 

[vela]

Homework Statement

Express $$ 2 arctan (\sqrt\frac{a-b}{a+b} tan (\theta/2))$$ in terms of inverse cosine

Homework Equations

I realize it amounts to find a smart substitution, but I can't find one.

The Attempt at a Solution

I tried $b/a=tan \theta$ , but I can't find any way to get rid of the other tangent term. I would be really nice if the solution is not a direct substitution of the enormous formula for the inverse trig functions, as it only gets messy. Thanks in advance!

Hint: Let $x = 2 \arctan \left[\sqrt\frac{a-b}{a+b} \tan \left(\frac \theta 2\right)\right]$. Then $\tan \frac x2 = \sqrt\frac{a-b}{a+b} \tan \left(\frac \theta 2\right)$. You can express $\tan\frac x2$ in terms of $\cos x$.

 

[cr7einstein]

Done! Thanks a lot @vela !