Official solution
S = \[{\tan ^{ - 1}}\left( {{1 \over 3}} \right) + {\tan ^{ - 1}}\left( {{1 \over 7}} \right) + {\tan ^{ - 1}}\left( {{1 \over {13}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {21}}} \right) + ....\]
= \[{\tan ^{ - 1}}\left( {{1 \over {1 + 1 \times 2}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {1 + 2 \times 3}}} \right) + ...\]
\[ \therefore \] Tr = \[{\tan ^{ - 1}}\left( {{1 \over {1 + r \times \left( {r + 1} \right)}}} \right)\]
= tan--1(r + 1) -- tan--1r
\[ \therefore \] T1
= tan--12 -- tan--11
T2
= tan--13 -- tan--12
T3
= tan--14 -- tan--13
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.
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T10 = tan-111 -- tan--110
\[ \therefore \] S = tan--111 -- tan--11 = \[{\tan ^{ - 1}}\left( {{{11 - 1} \over {1 + 11}}} \right)\]
\[ \therefore \] tan(S) = \[\tan \left( {{{\tan }^{ - 1}}\left( {{{11 - 1} \over {1 + 11}}} \right)} \right)\]
= \[{{{11 - 1} \over {1 + 11}}}\] = \[{{10} \over {12}} = {5 \over 6}\]