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Question Number 151665 by Huy last updated on 22/Aug/21

$$\mathrm{prove}4\mathrm{arccot}5-\mathrm{arccot}239=\frac{\pi }{4}$$

Answered by puissant last updated on 22/Aug/21

$$tan\left(4x\right)=\frac{4tanx-4{\left(tanx\right)}^3}{1-6{\left(tanx\right)}^2+{\left(tanx\right)}^4}$$$$tan\left(4arctan\frac{1}{5}\right)=\frac{120}{119}$$$$\beta =4tan\left(\frac{1}{5}\right)-arctan\left(\frac{1}{239}\right)$$$$\Rightarrow tan\beta =\frac{tan\left(4arctan\left(\frac{1}{5}\right)\right)-tan\left(arctan\left(\frac{1}{239}\right)\right)}{1+tan\left(4arctan\left(\frac{1}{5}\right)\right)tan\left(arctan\left(\frac{1}{239}\right)\right)}$$$$\Rightarrow tan\beta =\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119}\times \frac{1}{239}}=1$$$$\Rightarrow tan\beta =1\Rightarrow \beta =arctan\left(1\right)=\frac{\pi }{4}$$$$\because 4arctan\left(\frac{1}{5}\right)-arctan\left(\frac{1}{239}\right)=\frac{\pi }{4}$$$$\left(MACHINformula\right)..$$

Answered by qaz last updated on 22/Aug/21

$$\because \frac{{(5+\mathrm{i})}^4}{239+\mathrm{i}}=\frac{28561+28561\mathrm{i}}{60}$$$$\therefore 4\mathrm{arccot}5-\mathrm{arccot}239$$$$=4\arctan \frac{1}{5}-\arctan \frac{1}{239}$$$$=\arctan \left(\frac{\frac{28561}{60}}{\frac{28561}{60}}\right)$$$$=\frac{\pi }{4}$$

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