Explanation for an equality involving complexed exponents

Question

$$\frac{e^{i(N+1)x}-e^{-iNx}}{e^{ix}-1} = \frac{e^{i(N+1/2)x}-e^{-i(N+1/2)x}}{e^{ix/2}-e^{-ix/2}}$$

I'd be glad to get an explanation for both numerator and denominator.

Thanks in advance!

Answer 1

$\frac{e^{i(N+1)x}-e^{-iNx}}{e^{ix}-1} = \frac{(e^{i(N+1)x}-e^{-iNx})e^{-ix/2}}{(e^{ix}-1)e^{-ix/2}} = \frac{e^{i(N+1/2)x}-e^{-i(N+1/2)x}}{e^{ix/2}-e^{-ix/2}}$

Answer 2

Hint: $e^{ix} - 1= e^{ix/2}(e^{ix/2}-e^{-ix/2})$

Answer 3

Hint: Multiply with $\frac{e^{-ix/2}}{e^{-ix/2}}$.