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Question Number 77959 by john santu last updated on 12/Jan/20

solve tan ((1/(1+x^2 )))>1

$${solve}\:\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)>\mathrm{1}\: \\ $$

Answered by mr W last updated on 12/Jan/20

0<(1/(1+x^2 ))<1  nπ+(π/4)<(1/(1+x^2 ))<nπ+(π/2)  ⇒n=0  ⇒(π/4)<(1/(1+x^2 ))  ⇒x^2 <(4/π)−1  ⇒−(√((4/π)−1))<x<(√((4/π)−1))

$$\mathrm{0}<\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }<\mathrm{1} \\ $$ $${n}\pi+\frac{\pi}{\mathrm{4}}<\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }<{n}\pi+\frac{\pi}{\mathrm{2}} \\ $$ $$\Rightarrow{n}=\mathrm{0} \\ $$ $$\Rightarrow\frac{\pi}{\mathrm{4}}<\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$ $$\Rightarrow{x}^{\mathrm{2}} <\frac{\mathrm{4}}{\pi}−\mathrm{1} \\ $$ $$\Rightarrow−\sqrt{\frac{\mathrm{4}}{\pi}−\mathrm{1}}<{x}<\sqrt{\frac{\mathrm{4}}{\pi}−\mathrm{1}} \\ $$

Commented byjohn santu last updated on 12/Jan/20

thank you

$${thank}\:{you} \\ $$

Answered by MJS last updated on 12/Jan/20

tan α >1 ⇒ (π/4)+nπ<α<(π/2)+nπ; n∈Z  (π/4)+nπ<(1/(x^2 +1))<(π/2)+nπ  (2/((2n+1)π))−1<x^2 <(4/((4n+1)π))−1  the upper border must be greater than zero  (4/((4n+1)π))−1>0 ⇒ n=0  but (2/((2n+1)π))−1<0∧x^2 ≥0 ⇒  0≤x^2 <(4/π)−1  ⇒ −(√((4/π)−1))<x<+(√((4/π)−1))

$$\mathrm{tan}\:\alpha\:>\mathrm{1}\:\Rightarrow\:\frac{\pi}{\mathrm{4}}+{n}\pi<\alpha<\frac{\pi}{\mathrm{2}}+{n}\pi;\:{n}\in\mathbb{Z} \\ $$ $$\frac{\pi}{\mathrm{4}}+{n}\pi<\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}<\frac{\pi}{\mathrm{2}}+{n}\pi \\ $$ $$\frac{\mathrm{2}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}−\mathrm{1}<{x}^{\mathrm{2}} <\frac{\mathrm{4}}{\left(\mathrm{4}{n}+\mathrm{1}\right)\pi}−\mathrm{1} \\ $$ $$\mathrm{the}\:\mathrm{upper}\:\mathrm{border}\:\mathrm{must}\:\mathrm{be}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{zero} \\ $$ $$\frac{\mathrm{4}}{\left(\mathrm{4}{n}+\mathrm{1}\right)\pi}−\mathrm{1}>\mathrm{0}\:\Rightarrow\:{n}=\mathrm{0} \\ $$ $$\mathrm{but}\:\frac{\mathrm{2}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}−\mathrm{1}<\mathrm{0}\wedge{x}^{\mathrm{2}} \geqslant\mathrm{0}\:\Rightarrow \\ $$ $$\mathrm{0}\leqslant{x}^{\mathrm{2}} <\frac{\mathrm{4}}{\pi}−\mathrm{1} \\ $$ $$\Rightarrow\:−\sqrt{\frac{\mathrm{4}}{\pi}−\mathrm{1}}<{x}<+\sqrt{\frac{\mathrm{4}}{\pi}−\mathrm{1}} \\ $$

Commented byjohn santu last updated on 12/Jan/20

thank you

$${thank}\:{you} \\ $$

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