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Question Number 148221 by 0731619 last updated on 26/Jul/21

Answered by Olaf_Thorendsen last updated on 26/Jul/21

$$\mathrm{N}\left(x\right)={\tan }^2\left(\tan x\right)-{\tan }^{2}x$$$$\mathrm{N}\left(x\right)\underset{0}{\sim }{\tan }^2(x+\frac{x^3}{3}+\frac{2x^5}{15})-{(x+\frac{x^3}{3}+\frac{2x^5}{15})}^2$$$$\mathrm{N}\left(x\right)\underset{0}{\sim }((x+\frac{x^3}{3}+\frac{2x^5}{15})+\frac{1}{3}{(x+\frac{x^3}{3}+\frac{2x^5}{15})}^3$$$${+\frac{2}{15}{(x+\frac{x^3}{3}+\frac{2x^5}{15})}^5)}^2-{(x+\frac{x^3}{3}+\frac{2x^5}{15})}^2$$$$\mathrm{N}\left(x\right)\underset{0}{\sim }{((x+\frac{x^3}{3}+\frac{2x^5}{15})+(\frac{x^3}{3}+\frac{x^5}{3})+\frac{2}{15}{x}^5)}^2$$$$-(x^2+\frac{2x^4}{3}+\frac{17x^6}{45})$$$$\mathrm{N}\left(x\right)\underset{0}{\sim }{(x+\frac{2x^3}{3}+\frac{9x^5}{15})}^2-(x^2+\frac{2x^4}{3}+\frac{17x^6}{45})$$$$\mathrm{N}\left(x\right)\underset{0}{\sim }(x^2+\frac{4x^4}{3}+\frac{74x^6}{45})-(x^2+\frac{2x^4}{3}+\frac{17x^6}{45})$$$$\mathrm{N}\left(x\right)\underset{0}{\sim }\frac{2x^4}{3}+\frac{19x^6}{15}$$$$\Rightarrow p=9\times \frac{19}{15},n=4,q=9\times \frac{2}{3},r=0,t=0$$$$p=\frac{57}{5},q=6,r=0,n=4,t=0$$

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