tanya683
tanya683 tanya683
  • 25-06-2019
  • Mathematics
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jdoe0001 jdoe0001
  • 25-06-2019

[tex]\bf tan^2(x)-1=0\implies tan^2(x)=1\implies tan(x)=\pm\sqrt{1} \\\\\\ tan^{-1}[tan(x)]=tan^{-1}(\pm 1)\implies x=tan^{-1}(\pm 1)\implies x= \begin{cases} \frac{\pi }{4}\\\\ \frac{3\pi }{4}\\\\ \frac{5\pi }{4}\\\\ \frac{7\pi }{4} \end{cases}[/tex]

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