Question

If the value of $\left(1 + tan \, 1^{o}\right)\left(1 + tan \left( \, 2\right)^{o}\right)\left(1 + tan \, 3^{o}\right)$ $\ldots \ldots \ldots .\left(\right.1+tan\left( \, 44\right)^{o}\left.\right)\left(\right.1+tan \, \left(45\right)^{o}\left.\right)$ 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, $A+B=45^{o}$
$tan \left(A + B\right) = 1$
$\because tan \left(\right. A + B \left.\right) = \frac{tan A + tan B}{1 - tan A tan B}$
$\Rightarrow tan A+tan⁡B=1-tan⁡Atan⁡B$
$\Rightarrow \left(1 + \text{tan} A\right) \left(\right. 1 + \text{tan} B \left.\right) \, = \, 2$
LHS $= \left[\left(1 + tan 1^{o} \left.\right) \, \left(\right. 1 + tan \left(44\right)^{o} \left.\right)\right] \, \right[ \left(\right. 1 + tan 2^{o} \left.\right) \left(1 + tan \left(43\right)^{o} \left.\right)\right] \ldots \left[\right. \left(1 + tan \left(45\right)^{o} \left.\right)\right] \left[\text{for each} \left(1 + tan \theta \right) \left[1 + tan \left(\frac{\pi }{4} - \theta \right) = 2\right]\right]$
$= 2^{22} \, \left(1 + 1\right)$
$=2^{23}$
$=2^{\lambda }$
then, $\lambda =23.$
Hence the sum of digits of $\lambda $ is $2+3$
$=5$