How could I prove this trigonometric identity?

Question

Show that:

$$\left(\frac{1+\tan \theta}{1 - \tan \theta}\right)^n = \frac{1+i\tan n\theta}{1-i\tan n\theta}$$

Original image: https://i.stack.imgur.com/q8Yxj.jpg

Answer 1

The identity is false: for $\theta=\pi/3$ and $n=1$, the left hand side is $$ \frac{1+\sqrt{3}}{1-\sqrt{3}}=\frac{(1+\sqrt{3})^2}{1-3}=-2-\sqrt{3} $$ while the right hand side is $$ \frac{1+i\sqrt{3}}{1-i\sqrt{3}}=\frac{(1+i\sqrt{3})^2}{1+3}=\frac{-1+i\sqrt{3}}{2} $$ The correct identity is $$ \left(\frac{1+i\tan \theta}{1 - i\tan \theta}\right)^n = \frac{1+i\tan n\theta}{1-i\tan n\theta} $$ that follows at once by expanding $$ \tan\theta=\frac{\sin\theta}{\cos\theta} $$ and recalling De Moivre's identity.

Answer 2

As already explained by egreg, the identity $$ \left(\frac{1+\tan \theta}{1 - \tan \theta}\right)^n = \frac{1+i\tan n\theta}{1-i\tan n\theta} $$ is false.

However, the identity $$ \left(\frac{1+i\tan \theta}{1 - i\tan \theta}\right)^n = \frac{1+i\tan n\theta}{1-i\tan n\theta} $$ IS valid.

Indeed, \begin{align} \left(\frac{1+i\tan \theta}{1 - i\tan \theta}\right)^n &=\left(\frac{\cos\theta+i\sin \theta}{\cos\theta - i\sin \theta}\right)^n =\frac{(\cos\theta+i\sin \theta)^n}{(\cos\theta-i\sin \theta)^n}=\frac{\cos n\theta+i\sin n\theta}{\cos n\theta-i\sin n\theta} \\= \frac{1+i\tan n\theta}{1-i\tan n\theta}. \end{align} We have used DeMoivre's formula: $$ (\cos\theta\pm i\sin\theta)^n=\cos n\theta \pm i\sin n\theta. $$