# Converting one inverse trig function to another

## Homework Statement

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

## Homework Equations

I realise 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!!

## Answers and Replies

@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
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## Homework Statement

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

## Homework Equations

I realise 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 !!!