If the value of (1 + tan 1 ^ o )(1 + tan 2 ^ o )(1 + tan 3 ^ o ) ....

If, $A + B = {45}^{o}$

$\tan \left(A + B\right) = 1$
$∵ \tan ( A + B ) = \dfrac{tanA + tanB}{1 - tanAtanB}$

$⇒ tan A + tan B = 1 - tan Atan B$

$⇒ \left(1 + \text\tanA\right) ( 1 + \text\tanB ) = 2$

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

$= {2}^{23}$

$= {2}^{λ}$

then, $\lambda = 23.$

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

$= 5$