Let, $ z\;=\; (1+itan \alpha )^{1+itan\beta} \; \; where \; Z $ is real (i.e no imaginary part)

Taking log on both sides,

$ \therefore log z \;=\; (1+itan\beta) \; log(1+itan \alpha ) \\ \; \\ \; \\ = (1+itan\beta) \bigg[ log \sqrt{1+tan^2 \alpha} + i tan^{-1} \Big( \dfrac{tan \alpha}{1} \Big) \bigg] \\ \; \\ \; \\ \ldots \Bigg\{ \because log\sqrt{x+iy} \;=\; log\sqrt{x^2+y^2} + i tan^{-1} \Big( \dfrac{y}{x} \Big) \Bigg\} \\ \; \\ \; \\ (1+itan\beta) \bigg[ log (sec\alpha) + i tan^{-1} \alpha \bigg] \; \; \; \ldots \{ \because 1+tan^2 \theta \;=\; sec^2 \theta \} \\ \; \\ \; \\ = log(sec\alpha) + i tan\beta \cdot log(sec \alpha) + i \alpha - \alpha tan \beta \; \; \; [ \because i^2 \;=\; -1 ] \\ \; \\ \; \\ \therefore log z \;=\; [ log(sec \alpha) - \alpha tan \beta ] + i[ \alpha + tan \beta \cdot log(sec \alpha) ] \; \; \ldots (i) \\ \; \\ \; \\ $

Comparing real and imaginary parts,

$ \therefore i[ \alpha + tan \beta \cdot log(sec \alpha) ] \;=\; 0 \\ \; \\ \; \\ \therefore \alpha \;=\; - tan \beta \cdot log(sec \alpha) \; \; \; \; \; \; \ldots (ii) \\ \; \\ \; \\ \therefore logz \;=\; log(sec \alpha) \;-\; \alpha (tan \beta) \; \; \; \; \; \; \ldots From \; (i) \\ \; \\ \; \\ = log(sec \alpha) + tan \beta \cdot log(sec \alpha) \cdot tan \beta \; \; \; \; \; \ldots from \; (ii) \\ \; \\ \; \\ \therefore log z \;=\; log (sec \alpha) [1+tan^2 \beta] \\ \; \\ \; \\ = sec^2 \beta \cdot log(sec \alpha) \\ \; \\ \; \\ \therefore log z \;=\; log(sec \alpha)^{sec^2\beta} \\ \; \\ \; \\ \therefore z \;=\; (sec \alpha)^{sec^2\beta} \; \; \; \; \; \; \; \; \; \ldots Hence \; Proved. $