Suppose power ${\mathrm{P}}_{\mathrm{small}}$ is applied to the small gear such that:

${\mathrm{P}}_{\mathrm{small}}={\mathrm{T}}_{\mathrm{small}} {\mathrm{\omega }}_{\mathrm{small}}$

From earlier equations, the torque on the large gear can be expressed as:

${\mathrm{T}}_{\mathrm{large}}=\left(\frac{{\mathrm{D}}_{\mathrm{large}}}{{\mathrm{D}}_{\mathrm{small}}}\right){\mathrm{T}}_{\mathrm{small}}$

and the rotation speed magnitude can be expressed:

${\mathrm{\omega }}_{\mathrm{large}}=\left(\frac{{\mathrm{D}}_{\mathrm{small}}}{{\mathrm{D}}_{\mathrm{large}}}\right) {\mathrm{\omega }}_{\mathrm{small}}$

The power transmitted by the large gear is therefore:

$$\begin{gathered} {\mathrm{P}}_{\mathrm{large}}={\mathrm{T}}_{\mathrm{large}} {\mathrm{\omega }}_{\mathrm{large}} \\ =\left(\frac{{\mathrm{D}}_{\mathrm{large}}}{{\mathrm{D}}_{\mathrm{small}}}\right) {\mathrm{T}}_{\mathrm{small}} \left(\frac{{\mathrm{D}}_{\mathrm{small}}}{{\mathrm{D}}_{\mathrm{large}}}\right) {\mathrm{\omega }}_{\mathrm{small}} \\ ={\mathrm{T}}_{\mathrm{small}} {\mathrm{\omega }}_{\mathrm{small}} \\ \therefore {\mathrm{P}}_{\mathrm{large}}={\mathrm{P}}_{\mathrm{small}} \end{gathered}$$

The power transmitted by each shaft will always be the same. The trade-off will be an exchange between torque and rotation speed.