complex analysis - De Moivre's Theorem and a related formula?

$$(\cos(\theta)+i\sin(\theta))^n=\cos(n\theta)+i\sin(n\theta), n \in \mathbb{Z}$$

is De Moivre's Theorem. It is useful in calculating integer angle trigonometric identities such as $\cot(4\theta)$ by taking the real part of $(\cos(\theta)+i\sin(\theta))^4$ divided by its imaginary part, both parts obtained using binomial expansion.

However, from another post on this site, I've also seen that you can calculate $\cot(4\theta)$ by taking the real part of $(1+i\tan(\theta))^4$ divided by its imaginary part, likewise both parts obtained using binomial expansion.

So does that mean $(1+i\tan(\theta))^n=1+i\tan(n\theta)$? What's the relation between this formula and De Moivre's theorem?

Does this also mean I can flip the formulae to give me $(\sin(\theta)+i\cos(\theta))^4$ or $(\tan(\theta)+i)^4$ and take the imaginary part divided by the real part this time to still get the same answer for $\cot(4\theta)$?

Answer

Perhaps this may clarify.

$$(1+i \tan \theta)^4 = \sec^4 \theta (\cos \theta + i \sin \theta)^4 = \sec^4\theta (\cos 4\theta + i \sin 4\theta)$$
where we have used a variant of deMoivre's theorem in the last step. This also means, as the ratio of real to complex parts of the RHS is $\cot 4\theta$, you can use binomial theorem on the LHS and find the same ratio there to get an equivalent expression.

It does not imply $(1+i \tan \theta)^4$ is the same as $1 + i \tan 4\theta$.