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Question Number 5121 by Yozzii last updated on 15/Apr/16

What solutions x∈R exist for the equation (tan^(−1) x)(cot^(−1) x)=n where n∈Z?

$${What}\:{solutions}\:{x}\in\mathbb{R}\:{exist}\:{for}\:{the}\: \\ $$$${equation}\:\left({tan}^{−\mathrm{1}} {x}\right)\left({cot}^{−\mathrm{1}} {x}\right)={n} \\ $$$${where}\:{n}\in\mathbb{Z}? \\ $$

Answered by prakash jain last updated on 16/Apr/16

f(x)=tan^(−1) xcot^(−1) x f ′(x)=((cot^(−1) x−tan^(−1) x)/(1+x^2 )) f ′(x)=0 at x=1,x=−1 f(x) max at x=1,x=−1 and it takes value (π^2 /(16)). The only possible value for n in question is 0. tan^(−1) xcot^(−1) x=0 at x=0. So x=0 is the only solution.

$${f}\left({x}\right)=\mathrm{tan}^{−\mathrm{1}} {x}\mathrm{cot}^{−\mathrm{1}} {x} \\ $$$${f}\:'\left({x}\right)=\frac{\mathrm{cot}^{−\mathrm{1}} {x}−\mathrm{tan}^{−\mathrm{1}} {x}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${f}\:'\left({x}\right)=\mathrm{0}\:{at}\:{x}=\mathrm{1},{x}=−\mathrm{1} \\ $$$${f}\left({x}\right)\:{max}\:{at}\:{x}=\mathrm{1},{x}=−\mathrm{1}\:\mathrm{and}\:\mathrm{it}\:\mathrm{takes}\:\mathrm{value}\:\frac{\pi^{\mathrm{2}} }{\mathrm{16}}. \\ $$$$\mathrm{The}\:\mathrm{only}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{for}\:{n}\:\mathrm{in}\:\mathrm{question}\:\mathrm{is}\:\mathrm{0}. \\ $$$$\mathrm{tan}^{−\mathrm{1}} {x}\mathrm{cot}^{−\mathrm{1}} {x}=\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{0}. \\ $$$$\mathrm{So}\:{x}=\mathrm{0}\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solution}. \\ $$

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