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Q.
If $a =\displaystyle\lim _{ n \rightarrow \infty} \sum_{ k =1}^{ n } \frac{2 n }{ n ^2+ k ^2}$ and $f(x)=\sqrt{\frac{1-\cos x}{1+\cos x}}, x \in(0,1)$, then :
$ a =\frac{1}{ n } \displaystyle\sum_{ k =1}^{ n } \frac{2}{1+\left(\frac{ k }{ n }\right)^2}=\int_0^1 \frac{2}{1+ x ^2} dx =\frac{\pi}{2} $
$ f ( x )=\tan \left(\frac{ x }{2}\right) ; x \in(0,1) $
$ f \left(\frac{\pi}{4}\right)=\sqrt{2}-1 $
$ f ^{\prime}\left(\frac{\pi}{4}\right)=\frac{1}{2} \sec ^2\left(\frac{\pi}{8}\right)=\frac{\sqrt{2}}{\sqrt{2}+1}$
$ f ^{\prime}\left(\frac{\pi}{4}\right)=\sqrt{2} f \left(\frac{\pi}{4}\right)$