# Complex Numbers Euler's Identity

I am trying to follow a solution to a problem and it says that $\left(\frac{1+i\tan\theta}{1-i\tan\theta}\right)^a = \left(\frac{\cos\theta+i\sin\theta}{\cos\theta-i\sin\theta}\right)^a$

However, I am unable to see why this would be? It is probably just a simple oversight but I am unsure.

Simply multiply both the numerator and denominator of the left hand side (within the parentheses) by $\cos\theta$ and note that $\tan\theta = \sin\theta / \cos\theta$

$$\left( \frac{\cos\theta\left( 1 + i\frac{\sin\theta}{\cos\theta} \right)}{\cos\theta\left( 1 - i\frac{\sin\theta}{\cos\theta} \right)} \right)^a$$

HINT:

Multiply the numerator and denominator of L.H.S. by $\cos ^a \theta$ and you will get R.H.S.

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

Since

$$\tan \theta = {{\sin \theta} \over {\cos \theta}}$$

You can cancel out the $a$ on both sides first, then, on the RHS, divide the top and bottom by $\cos \theta$.

That should leave the two being equal. Then you can raise to the power of $a$ again on both sides.

Treating $i$ as a constant , leaving exponent $a$ alone, using $\tan\theta = \sin\theta / \cos\theta,$ it comes about.