Answer
See explanation
Work Step by Step
The efficiency is defined as$$
\eta_{\propto, c}=\frac{d h_s}{d h}
$$For an ideal gas,
$$
d h=c_p d T
$$From Gibbs second equation,$$
d h=T d s+u d P
$$For the isentropic case,$$
d h_s=u d P
$$Substituting, we obtain$$
\eta_{a n, C}=\frac{u d P}{c_p d T}
$$Then,$$
\begin{aligned}
& \eta_{\infty, C} c_p d T=u d P \\
& \eta_{\infty, C} c_p d T=\frac{R T}{P}
\end{aligned}
$$Integrating between inlet (1) and exit (2) states,$$
\begin{aligned}
& \int_1^2 \frac{1}{\eta_{\infty}, C} \frac{R}{c_P} \frac{d P}{P}=\int_1^2 \frac{d T}{T} \\
& \frac{1}{\eta_{\infty}, C} \frac{R}{c_P} \ln \frac{P_2}{P_1}=\ln \frac{T_2}{T_1}
\end{aligned}
$$This becomes$$
\frac{T_2}{T_1}=\left(\frac{P_2}{P_1}\right)^{\frac{1}{\eta_{\infty, c} C_p}}=\left(\frac{P_2}{P_1}\right)^{\frac{1}{\eta_{\infty, c} \frac{k-1}{k}}}
$$which is the desired expression.
