Question

If the value of $\left(1+tan{1}^{\circ }\right)\left(1+tan{2}^{\circ }\right)\left(1+tan{3}^{\circ }\right)$ $\dots \dots \dots .(1+tan44^{\circ })(1+tan45^{\circ })$ 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, $A+B=45^{\circ }$
$tan\left(A+B\right)=1$
$\because tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}$
$\Rightarrow tanA+tanB=1-tanAtanB$
$\Rightarrow \left(1+tanA\right)(1+tanB)=2$
LHS
$=\left[\left(1+tan{1}^{\circ })(1+tan44^{\circ })\right]\right[(1+tan{2}^{\circ })\left(1+tan43^{\circ })\right]\dots [\left(1+tan45^{\circ })\right]$
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 $2+3$ $=5$