A year ahead!

By observation,

${ T }_{ r }=\sec { \frac { (2r-2)\pi }{ 2016 } } \sin { \frac { (2r-1)\pi }{ 2016 } } \sec { \frac { 2r\pi }{ 2016 } }$

Let $A=\frac { (2r-2)\pi }{ 2016 }$ and $B=\frac { 2r\pi }{ 2016 }$ .

Now,

${ T }_{ r }=\sec { A } \sin { \frac { A+B }{ 2 } } \sec { B }$

${ T }_{ r }=\frac { \sin { \frac { A+B }{ 2 } } }{ \cos { A } \cos { B } }$

Multiplying and dividing with $2\sin { \frac { A-B }{ 2 } }$ ,

${ T }_{ r }=\frac { 1 }{ 2\sin { \frac { A-B }{ 2 } } } \frac { 2\sin { \frac { A+B }{ 2 } } \sin { \frac { A-B }{ 2 } } }{ \cos { A } \cos { B } }$

Using trignometric identity,

${ T }_{ r }=\frac { 1 }{ 2\sin { \frac { A-B }{ 2 } } } \frac { \cos { B } -\cos { A } }{ \cos { A } \cos { B } }$

Multiplying and dividing by $\cos A\cos B$ ,

${ T }_{ r }=\frac { 1 }{ 2\sin { \frac { A-B }{ 2 } } } (\sec { A } -\sec { B } )$

Replacing $A$ and $B$ with $\frac { (2r-2)\pi }{ 2016 }$ and $\frac { 2r\pi }{ 2016 }$ respectively,

${ T }_{ r }=\frac { 1 }{ 2\sin { \frac { \pi }{ 2016 } } } (\sec { \frac { 2r\pi }{ 2016 } } -\sec { \frac { (2r-2)\pi }{ 2016 } } )$

We have been able to express the $rth$ term as difference of two consecutive terms.

$s=\sum _{ 1 }^{ 336 }{ \frac { 1 }{ 2\sin { \frac { \pi }{ 2016 } } } (\sec { \frac { 2r\pi }{ 2016 } } -\sec { \frac { (2r-2)\pi }{ 2016 } } ) }$

For small values of $x$ ,

${ \sin { x } }\approx x$ ,

$s=\sum _{ 1 }^{ 336 }{ \frac { 1008 }{ \pi } (\sec { \frac { 2r\pi }{ 2016 } } -\sec { \frac { (2r-2)\pi }{ 2016 } } ) }$

We observe that this is a telescoping series.

$336$ goes to termwise greater term and $1$ goes to termwise smaller term.

$s=\frac { 1008 }{ \pi } (\sec { \frac { 2(336)\pi }{ 2016 } } -\sec { \frac { (2(1)-2)\pi }{ 2016 } } )$

$s=\frac { 1008 }{ \pi } (\sec { \frac { \pi }{ 3 } } -\sec { 0 } )$

$s=\frac { 1008 }{ \pi } (2-1)$

$s=\frac { 1008 }{ \pi }$

$s\pi =1008$