Let  $z= (1+ i \tan \alpha ) ^{(1+i \tan \beta)}$

Taking logarithms on both sidex,

Hence,

$\begin {align} \log z &= (1+i \tan)\log (1+ i \tan \alpha) \\ &= (1+i \tan \beta) \left [ \dfrac {1}{2} \log (1+ \tan ^2 \alpha) + i \tan ^{-1}\alpha \right ] \end{align} $

$= (1+ i \tan \beta) \big [ \log( \sec \alpha ) + i \alpha \big]$

$\therefore \log z = (\log \sec \alpha - \alpha \tan \beta) + i (\alpha + \tan \beta \log \sec \alpha ) = x + iy $ where

$x= (\log \sec \alpha - \alpha \tan \beta)$ and $y=\alpha + \tan \beta \log \sec \alpha \ \cdots \ \cdots (1)$

Now, $z = e^{x+iy} = e^x e^{iy} = e^x (\cos y + i\sin y)$

Since by data, z is real, $e^x \sin y =0 \ \therefore y =0 \ \therefore \cos y=1$

$\therefore z=e^x \cos y =e^x =e ^{(\log \sec \alpha - \alpha \tan \beta)} = e^{(\log \sec \alpha )} e^{-\alpha \tan \beta}\ \cdots \ \cdots (2)$

But since $y=0,$ from (1),  $\alpha + \tan \beta \log\sec \alpha =0$

$\therefore -\alpha = \tan \beta \log \sec \alpha$

$\therefore - \alpha \tan \beta = \tan ^2 \beta \log \sec \alpha = \log (\sec \alpha ) ^{\tan^2 \beta}$

$\therefore e^{-\alpha \tan \beta} = (\sec \alpha) ^{\tan^2 \beta}$

$\therefore z = \sec \alpha (\sec \alpha) ^{\tan ^2 \beta} = (\sec \alpha ) ^{1+ \tan^2 \beta} = (\sec \alpha)^{\sec^2 \beta}$ (from 2)

$\therefore z= (\sec \alpha) ^{\sec ^2 \beta}$

Hence proved....