$Let \ x+iy = (1+i \tan \alpha)^{(1+i\tan \beta )}$ $\log (x+iy)=\log (1+ i \tan \alpha)^{(1+i\tan \alpha)}= (1+i\tan \beta) \log (1+ i \tan \alpha)$ $\log \sqrt{x^2+y^2 }+ i \tan^{-1} \left ( \dfrac{y}{x} \right ) = (1+ i \tan \beta) \left [ \log \sqrt{1^2 +tan^2 \alpha}+ i \tan^{-1} \left ( \dfrac {\tan \alpha}{1} \right ) \right ]$ $\log \sqrt{x^2 +y^2 }+ i \tan^{-1} \left ( \dfrac {y}{x} \right )= (1+i \tan \beta) [ \log \sec \alpha + i \alpha] \left { \because , 1+\tan^2 A=\sec^2 A \right }$ $\log \sqrt{x^2 +y^2 }+ i \tan^{-1} \left ( \dfrac {y}{x} \right )= (1+i \tan \beta) [ \log \sec \alpha + i \alpha] \left { \because , 1+\tan^2 A=\sec^2 A \right }$ $\because. (1+ i \tan \alpha)^{(1+i\tan \beta)} = x+ iy \ is \ real, \ y=0$ $\log x+i0= (\log \sec \alpha - \alpha \tan \beta ) + i (\alpha + \tan \beta \log \sec \alpha)$ Comparing real and imaginary parts, $\log x= (\log \sec \alpha - \alpha \tan \beta ) \ and \ 0 = (\alpha + \tan \beta \log \sec \alpha )$ $\therefore , \alpha = - \tan \beta \log \sec \alpha$ $\therefore, \log x = \log \sec \alpha + \tan^2 \beta \log sec \alpha = \log \sec \alpha (1+ \tan^2 \beta )$ $\log x = \log \sec \alpha \times \sec^2 \beta = \log \sec \alpha^{\sec^2 \beta }$ $\therefore , x= \sec \alpha^{\sec^2 \beta}$ $\therefore \ one \ of \ the \ principle \ values \ of \ (1+i \tan \alpha)^{(1+i \tan \beta )} = \sec\alpha^{\sec^2 \beta}$