Do we need to calculate each and every term

$\begin{aligned}{w}_{n+1}&=\frac{1+{w}_{n}}{1-{w}_{n}}\\&=\frac{1+\frac{1+{w}_{n-1}}{1-{w}_{n-1}}}{1-\frac{1+{w}_{n-1}}{1-{w}_{n-1}}}\\&=-\frac{1}{{w}_{n-1}}\\{w}_{2016}&=-\frac{1}{{w}_{2014}}\\&=-\frac{1}{-\frac{1}{{w}_{2012}}}\\&\huge\color{#3D99F6}{=\boxed{{w}_{2012}}}\end{aligned}$

Just observe a simple recurrence

w(0) = 2015

w(1)= -1008/1007

w(2) = -1/2015

w(3)= 1007/1008

w(4)=2015

Thus for every w(n) where n=4k

we have w(n) = 2015

Thus w(2016)=w(2012)

Let $w_{n}=\tan(x_{n})$ for some $x_{n} \in R$ and $n \in N$ .

This is because the range of $\tan x$ is R so any term of sequence(which consists of real terms) is equal to $\tan x$ for some $x \in R$

Then $\frac{1+w_{n}}{1-w_{n}}=\frac{1+\tan(x_{n})}{1-\tan(x_{n})}=\tan(\frac{\pi}{4}+x_{n})=\tan(x_{n+1})=w_{n+1}$

So we get

$w_{n+2}=\tan(x_{n+2})=\tan(\frac{\pi}{2}+x_{n})$

$w_{n+3}= \tan(x_{n+3})=\tan(\frac{3\pi}{4}+x_{n})$

$w_{n+4}=\tan(x_{n+4})=\tan(\pi+x_{n})=\tan(x_{n})=w_{n}$

We conclude that the sequence is periodic with period $n = 4$ and hence answer is $w_{2016}=w_{2012}$