Question

If the value of $\left(1 + tan \, 1 ^\circ \right)\left(1 + tan \, 2 ^\circ \right)\left(1 + tan \, 3 ^\circ \right)$ $\ldots \ldots \ldots .\left(\right.1+tan \, 44^\circ \left.\right)\left(\right.1+tan \, 45^\circ \left.\right)$ is $2^{\lambda }$ $,$ then the sum of the digits of the number $\lambda $ is

• A. $3$
• B. $6$
• C. $5$
• D. $4$

Answer 1

Answer: (C)

If, $A+B=45^\circ $
$tan\left(A + B\right)=1$
$\because tan \left(\right. A + B \left.\right) = \frac{tan A + tan B}{1 - tan A tan B}$
$\Rightarrow tanA+tanB=1-tanAtanB$
$\Rightarrow \left(1 + tan A\right)\left(\right.1+tanB\left.\right)=2$
LHS
$=\left[\left(1 + tan 1 ^\circ \left.\right) \left(\right. 1 + tan 44 ^\circ \left.\right)\right] \right[\left(\right.1+tan2^\circ \left.\right)\left(1 + tan 43 ^\circ \left.\right)\right]\ldots \left[\right.\left(1 + tan 45 ^\circ \left.\right)\right]$
for each $\left[\left(1 + tan \theta \right) \left[1 + tan \left(\frac{\pi }{4} - \theta \right) = 2\right]\right]$
$= 2^{22} \left(1 + 1\right)$
$=2^{23}$
$=2^{\lambda }$
Then, $\lambda =23.$
Hence, the sum of digits of $\lambda $ is $2+3$ $=5$