Happy New Year

$\begin{aligned} \text{It is give that:} \quad \tan \theta + \cot \theta & = 2 \\ \Rightarrow \tan \theta + \frac{1}{\tan \theta} & = 2 \\ \tan^2 \theta - 2 \tan \theta + 1 & = 0 \\ (\tan \theta - 1)^2 & = 0 \\ \Rightarrow \tan \theta & = 1 \\ \cot \theta & = 1 \end{aligned}$

Therefore,

$\begin{aligned} 2^A & = \sum_{n=1}^{2^{2016}} \tan^{2^n} \theta + \cot^{2^n} \theta = \sum_{n=1}^{2^{2016}} 2 = 2^{2016}(2) = 2^{2017} \\ \Rightarrow A & = \boxed{2017} \end{aligned}$

tanθ+ $1/tanθ$ ≥2 ∀tanθ>0 and equality occurs(which is the case in the question) when tanθ=1. Therefore summation simplifies to \displaystyle \sum_{i=1}^2^{2016} (2) = $2.2^{2016} = 2^{2017}$

It is given that : tanθ +cotθ =2

      square it on both sides  to get 
                                    tan^2 (θ )+cot2θ = 2

     square it again   and go on,
                              you will get a generalised equation

                                    tan^2^n (θ) +cot^2^n (θ) = 2
    final answer =   (2)  2^2016 =2^2017