Inverse Trig Function

I have reached a part of an example physics problem in which they go from one line to the next and I cannot figure out how.

So my task is to show that

$$\tan^{-1}\frac{sin\theta}{\cos\theta - 1} = 90^{\circ} + \frac{\theta}{2}$$

Can I get a hint in the right direction?

Thanks!

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You know that Tan(90+theta) = Tan(theta), so the 90 goes away. The rest should be a half-angle identity.

You know that Tan(90+theta) = Tan(theta), so the 90 goes away.
That's not true. $\tan(x+\pi) = \tan x$, but $\tan(x+\pi/2)\not=\tan x$.

Mark44
Mentor
I have reached a part of an example physics problem in which they go from one line to the next and I cannot figure out how.

So my task is to show that

$$\tan^{-1}\frac{sin\theta}{\cos\theta - 1} = 90^{\circ} + \frac{\theta}{2}$$

Can I get a hint in the right direction?

Thanks!
Your equation is equivalent to $$\frac{sin\theta}{\cos\theta - 1}~=~tan(90^{\circ} + \frac{\theta}{2})~=~tan(1/2(\theta + \pi))$$

You can use one of the half-angle formulas for tangent to work with the expression on the right.