Question-146653

Question Number 146653 by mathdanisur last updated on 14/Jul/21

Answered by mindispower last updated on 14/Jul/21

tg((π/2)−x)=(1/(tg(x))) S=Σ_(k=20) ^(70) ((tg(k))/(1+tg(k))) =Σ_(k=20) ^(44) ((tg(k))/(1+tg(k)))+((tg(45))/(tg(45)+1))+Σ_(k=46) ^(70) ((tg(k))/(1+tg(k))) =Σ_(k=20) ^(44) ((tg(k))/(1+tg(k)))+((1 )/2)+Σ_(20) ^(44) ((tg(90−k))/(1+tg(90−k))) =Σ_(k=20) ^(44) (((tg(k))/(1+tg(k)))+((1/(tg(k)))/(1+(1/(tg(k))))))+(1/2) =Σ_(k=20) ^(44) ((1+tg(k))/(1+tg(k)))+(1/2)=Σ_(k=20) ^(44) 1+(1/2)=((51)/2)=25,5

$${tg}\left(\frac{\pi}{\mathrm{2}}−{x}\right)=\frac{\mathrm{1}}{{tg}\left({x}\right)} \\ $$$${S}=\underset{{k}=\mathrm{20}} {\overset{\mathrm{70}} {\sum}}\frac{{tg}\left({k}\right)}{\mathrm{1}+{tg}\left({k}\right)} \\ $$$$=\underset{{k}=\mathrm{20}} {\overset{\mathrm{44}} {\sum}}\frac{{tg}\left({k}\right)}{\mathrm{1}+{tg}\left({k}\right)}+\frac{{tg}\left(\mathrm{45}\right)}{{tg}\left(\mathrm{45}\right)+\mathrm{1}}+\underset{{k}=\mathrm{46}} {\overset{\mathrm{70}} {\sum}}\frac{{tg}\left({k}\right)}{\mathrm{1}+{tg}\left({k}\right)} \\ $$$$=\underset{{k}=\mathrm{20}} {\overset{\mathrm{44}} {\sum}}\frac{{tg}\left({k}\right)}{\mathrm{1}+{tg}\left({k}\right)}+\frac{\mathrm{1}\:}{\mathrm{2}}+\underset{\mathrm{20}} {\overset{\mathrm{44}} {\sum}}\frac{{tg}\left(\mathrm{90}−{k}\right)}{\mathrm{1}+{tg}\left(\mathrm{90}−{k}\right)} \\ $$$$=\underset{{k}=\mathrm{20}} {\overset{\mathrm{44}} {\sum}}\left(\frac{{tg}\left({k}\right)}{\mathrm{1}+{tg}\left({k}\right)}+\frac{\frac{\mathrm{1}}{{tg}\left({k}\right)}}{\mathrm{1}+\frac{\mathrm{1}}{{tg}\left({k}\right)}}\right)+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\underset{{k}=\mathrm{20}} {\overset{\mathrm{44}} {\sum}}\frac{\mathrm{1}+{tg}\left({k}\right)}{\mathrm{1}+{tg}\left({k}\right)}+\frac{\mathrm{1}}{\mathrm{2}}=\underset{{k}=\mathrm{20}} {\overset{\mathrm{44}} {\sum}}\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{51}}{\mathrm{2}}=\mathrm{25},\mathrm{5} \\ $$$$ \\ $$$$ \\ $$

Commented by Ar Brandon last updated on 14/Jul/21