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

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

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

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

Commented byjohn santu last updated on 12/Jan/20

$$thankyou$$

Answered by MJS last updated on 12/Jan/20

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

Commented byjohn santu last updated on 12/Jan/20

$$thankyou$$

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