Question Number 146653 by mathdanisur last updated on 14/Jul/21
Answered by mindispower last updated on 14/Jul/21
$${tg}\left(\frac{\pi}{\mathrm{2}}−{x}\right)=\frac{\mathrm{1}}{{tg}\left({x}\right)} \\ $$$${S}=\underset{{k}=\mathrm{20}} {\overset{\mathrm{70}} {\sum}}\frac{{tg}\left({k}\right)}{\mathrm{1}+{tg}\left({k}\right)} \\ $$$$=\underset{{k}=\mathrm{20}} {\overset{\mathrm{44}} {\sum}}\frac{{tg}\left({k}\right)}{\mathrm{1}+{tg}\left({k}\right)}+\frac{{tg}\left(\mathrm{45}\right)}{{tg}\left(\mathrm{45}\right)+\mathrm{1}}+\underset{{k}=\mathrm{46}} {\overset{\mathrm{70}} {\sum}}\frac{{tg}\left({k}\right)}{\mathrm{1}+{tg}\left({k}\right)} \\ $$$$=\underset{{k}=\mathrm{20}} {\overset{\mathrm{44}} {\sum}}\frac{{tg}\left({k}\right)}{\mathrm{1}+{tg}\left({k}\right)}+\frac{\mathrm{1}\:}{\mathrm{2}}+\underset{\mathrm{20}} {\overset{\mathrm{44}} {\sum}}\frac{{tg}\left(\mathrm{90}−{k}\right)}{\mathrm{1}+{tg}\left(\mathrm{90}−{k}\right)} \\ $$$$=\underset{{k}=\mathrm{20}} {\overset{\mathrm{44}} {\sum}}\left(\frac{{tg}\left({k}\right)}{\mathrm{1}+{tg}\left({k}\right)}+\frac{\frac{\mathrm{1}}{{tg}\left({k}\right)}}{\mathrm{1}+\frac{\mathrm{1}}{{tg}\left({k}\right)}}\right)+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\underset{{k}=\mathrm{20}} {\overset{\mathrm{44}} {\sum}}\frac{\mathrm{1}+{tg}\left({k}\right)}{\mathrm{1}+{tg}\left({k}\right)}+\frac{\mathrm{1}}{\mathrm{2}}=\underset{{k}=\mathrm{20}} {\overset{\mathrm{44}} {\sum}}\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{51}}{\mathrm{2}}=\mathrm{25},\mathrm{5} \\ $$$$ \\ $$$$ \\ $$
Commented by Ar Brandon last updated on 14/Jul/21
$${Bonsoir}\:{monsieur}\:! \\ $$
Commented by mindispower last updated on 14/Jul/21
$${bonsoir}\:{vous}\:{allez}\:{bien} \\ $$
Commented by mathdanisur last updated on 14/Jul/21
$${alot}\:{cool}\:{Ser}\:{thankyou} \\ $$
Commented by Ar Brandon last updated on 14/Jul/21
Merci. vous de même.