<s>[INST] Separate Into Real and Imaginary Parts Separate the following trigonometric function into Real and Imaginary Parts 
$$\tan^{-1}e^{i\theta} $$
or
$$\tan^{-1}(\cos\theta+i\sin\theta)$$
I Have made till here
Assuming $x+iy$ is the final outcome after separation
$$\therefore \tan^{-1}e^{i\theta}=x+iy$$
$$\therefore e^{i\theta}=\tan(x+iy)$$
$$\therefore e^{i\theta}= \frac{\sin2x}{cos2x+cos2hy}+i \frac{\sin2hy}{cos2x+cos2hy}  $$
$$\therefore \cos\theta+i\sin\theta= \frac{\sin2x}{cos2x+cos2hy}+i \frac{\sin2hy}{cos2x+cos2hy}  $$
But How to proceed from here to find the value of $ x $ and $ y $ so as to get $x+iy$
Or is there any other way to get the answer other than the steps that I have used.
 [/INST] I have come up with this solution Hope this might be the correct one
Let$$\displaystyle\tan^{-1}e^{i\theta}=x+iy$$
$$e^{i\theta}=\tan(x+iy)$$
$$\cos\theta+i\sin\theta=\tan(x+iy)$$
$$\cos\theta-i\sin\theta=tan(x-iy)$$
Now,$$ \tan2x=tan[(x+iy)+(x-iy)]$$
$$=\frac{\tan(x+iy)+\tan(x-iy)}{1-\tan(x+iy)\tan(x-iy)}$$
$$=\frac{(\cos\theta+i\sin\theta)+(\cos\theta-isin\theta)}{1-(\cos\theta+isin\theta)(\cos\theta-isin\theta)}$$
$$=\frac{2\cos\theta}{1-(\cos^2\theta+isin^2\theta)}=\frac{2\cos\theta}{1-1}=\frac{2cos\theta}{0}=\infty$$
$$\therefore 2x=\frac{\pi}{2}\therefore x=\frac{\pi}{4}$$
Also,
$$ \tan2iy=tan[(x+iy)-(x-iy)]$$
$$=\frac{\tan(x+iy)-\tan(x-iy)}{1+\tan(x+iy)\tan(x-iy)}$$
$$=\frac{(\cos\theta+i\sin\theta)-(\cos\theta-isin\theta)}{1+(\cos\theta+isin\theta)(\cos\theta-isin\theta)}$$
$$=\frac{2i\sin\theta}{1+(\cos^2\theta+isin^2\theta)}=\frac{2i\sin\theta}{2}$$
$$\therefore i\tanh2y=i\sin\theta\therefore \tanh2y=\sin\theta$$
$$\therefore 2y=\tanh^{-1}\sin\theta\therefore y=\frac{1}{2}\tanh^{-1}\sin\theta$$
Real part x=$\frac{\pi}{4}$,
Imaginary part y=$\frac{1}{2}\tanh^{-1}\sin\theta$
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