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Question Number 156807 by amin96 last updated on 15/Oct/21

f(x)=arctg(1/(x^2 +x+1)) and α=f(1)+f(2)+…+f(21) find tg(α)=?

$${f}\left({x}\right)={arctg}\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:\:{and}\:\alpha={f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+\ldots+{f}\left(\mathrm{21}\right) \\ $$$${find}\:\:{tg}\left(\alpha\right)=? \\ $$

Commented by amin96 last updated on 15/Oct/21

A)(6/(11)) B)(7/(11)) C)((11)/(21)) D)(1/(21)) E)((21)/(23))

$$\left.{A}\left.\right)\left.\frac{\mathrm{6}}{\mathrm{11}}\left.\:\left.\:\:\:{B}\right)\frac{\mathrm{7}}{\mathrm{11}}\:\:\:{C}\right)\frac{\mathrm{11}}{\mathrm{21}}\:\:\:\:{D}\right)\frac{\mathrm{1}}{\mathrm{21}}\:\:\:\:{E}\right)\frac{\mathrm{21}}{\mathrm{23}} \\ $$

Commented by amin96 last updated on 15/Oct/21

nope sir

$${nope}\:{sir} \\ $$

Commented by ghimisi last updated on 15/Oct/21

arctg(1/(x^2 +x+1))=arctg(1/x)−arctg(1/(x+1))

$${arctg}\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}={arctg}\frac{\mathrm{1}}{{x}}−{arctg}\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$

Answered by gsk2684 last updated on 16/Oct/21

f(x)=tan^(−1) (1/(1+(x+1)x))=tan^(−1) (((x+1)−x)/(1+(x+1)x))=tan^(−1) (x+1)−tan^(−1) x α=f(1)+f(2)+f(3)+....+f(21) α=(tan^(−1) 2−tan^(−1) 1) +(tan^(−1) 3−tan^(−1) 2) +(tan^(−1) 4−tan^(−1) 3) +... +(tan^(−1) 22−tan^(−1) 21) α=tan^(−1) 22−tan^(−1) 1=tan^(−1) ((22−1)/(1+22×1))=tan^(−1) ((21)/(23))

$${f}\left({x}\right)=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+\left({x}+\mathrm{1}\right){x}}=\mathrm{tan}^{−\mathrm{1}} \frac{\left({x}+\mathrm{1}\right)−{x}}{\mathrm{1}+\left({x}+\mathrm{1}\right){x}}=\mathrm{tan}^{−\mathrm{1}} \left({x}+\mathrm{1}\right)−\mathrm{tan}^{−\mathrm{1}} {x} \\ $$$$\alpha={f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+{f}\left(\mathrm{3}\right)+....+{f}\left(\mathrm{21}\right) \\ $$$$\alpha=\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{2}−\mathrm{tan}^{−\mathrm{1}} \mathrm{1}\right) \\ $$$$\:\:\:\:+\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{3}−\mathrm{tan}^{−\mathrm{1}} \mathrm{2}\right) \\ $$$$\:\:\:\:+\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{4}−\mathrm{tan}^{−\mathrm{1}} \mathrm{3}\right) \\ $$$$\:\:\:\:+... \\ $$$$\:\:\:\:+\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{22}−\mathrm{tan}^{−\mathrm{1}} \mathrm{21}\right) \\ $$$$\alpha=\mathrm{tan}^{−\mathrm{1}} \mathrm{22}−\mathrm{tan}^{−\mathrm{1}} \mathrm{1}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{22}−\mathrm{1}}{\mathrm{1}+\mathrm{22}×\mathrm{1}}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{21}}{\mathrm{23}} \\ $$

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