Showing $\frac{2\sin\theta\cos i\phi}{2\cos\theta\sin i\phi}=\frac{\tan\theta}{i\tanh\phi}$

Question

I have encountered a problem in an undergraduate Bsc course text book as follows.

$$\frac{2\sin\theta\cos i\phi}{2\cos\theta\sin i\phi}=\frac{\tan\theta}{i\tanh\phi}$$

I think this relation is not true. Please enlighten me. I tried by putting the trigonometric functions in their corresponding exponential forms, but the relation couldn't be established that way.

Answer 1

As noted in the comments, the proof is straightforward.

$\frac{2\sin\theta\cos i\phi}{2\cos\theta\sin i\phi}$

$=\frac{\sin\theta\cosh\phi}{\cos\theta i\sinh\phi}$

$=\tan\theta\times\frac{\coth\phi}{i}$

$=\tan\theta\times\frac{1}{i\tanh\phi}$

$=\frac{\tan\theta}{i\tanh\phi}$.