If $$a = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{{2n} \over {{n^2} + {k^2}}}} $$ and $f(x) = \sqrt {{{1 - \cos x} \over {1 + \cos x}}}$, $x \in (0,1)$, then :

<p>$$a = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{{2n} \over {{n^2} + {k^2}}}} $$</p> <p>$$ = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{k = 1}^n {{2 \over {1 + {{\left( {{k \over n}} \right)}^2}}}} $$</p> <p>$$a = \int\limits_0^1 {{2 \over {1 + {x^2}}}dx = 2{{\tan }^{ - 1}}x\int_0^1 { = {\pi \over 2}} } $$</p> <p>$f(x) = \sqrt {{{1 - \cos x} \over {1 + \cos x}}} ,\,x \in (0,1)$</p> <p>$f(x) = {{1 - \cos x} \over {\sin x}} = \cos ec\,x - \cot x$</p> <p>$f'(x) = \cos e{c^2}x - \cos ec\,x\cot x$</p> <p>$$\left. {\matrix{ {f\left( {{a \over 2}} \right) = f\left( {{\pi \over 4}} \right) = \sqrt 2 - 1} \cr {f'\left( {{a \over 2}} \right) = f'\left( {{\pi \over 4}} \right) = 2 - \sqrt 2 } \cr } } \right\}f'\left( {{a \over 2}} \right) = \sqrt 2 \,.\,f\left( {{a \over 2}} \right)$$</p>