If, $A+B={45}^{\mathrm{o}}$

$\tan \left(A+B\right)=1$
$\because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

$\Rightarrow \tan A+\tan B=1-\tan A\tan B$

$\Rightarrow \left(1+\text{tan}A\right)(1+\text{tan}B) = 2$

LHS$$=\left[\left(1+\tan 1^o) (1+\tan {44}^o)\right] \right[(1+\tan 2^o)\left(1+\tan {43}^o)\right]\dots [\left(1+\tan {45}^o)\right]\left[\text{for each}\left(1+\tan \theta \right)\left[1+\tan \left(\frac{\pi }{4}-\theta \right)=2\right]\right]$$
$=2^{22} \left(1+1\right)$

$=2^{23}$

$=2^{\lambda }$

then, $\lambda =23.$ 

Hence the sum of digits of $\lambda$ is $2+3$

$=5$