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

$$f\left(x\right)=arctg\frac{1}{x^2+x+1}and\alpha =f\left(1\right)+f\left(2\right)+\dots +f\left(21\right)$$$$findtg\left(\alpha \right)=?$$

Commented by amin96 last updated on 15/Oct/21

$$A)\frac{6}{11}B)\frac{7}{11}C)\frac{11}{21}D)\frac{1}{21}E)\frac{21}{23}$$

Commented by amin96 last updated on 15/Oct/21

$$nopesir$$

Commented by ghimisi last updated on 15/Oct/21

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

Answered by gsk2684 last updated on 16/Oct/21

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

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