Question

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