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Question Number 31734 by mondodotto@gmail.com last updated on 13/Mar/18

Answered by Tinkutara last updated on 13/Mar/18

$$4t-4t^3=1-6t^2+t^4$$$$\frac{4t-4t^3}{1-6t^2+t^4}=1$$$$Ift=\tan \theta ,then\tan 4\theta =1=\tan \frac{\pi }{4}$$$$4\theta =n\pi +\frac{\pi }{4}$$$$\theta =\frac{(4n+1)\pi }{16}$$$$t=\tan \frac{(4n+1)\pi }{16},n\in Z$$

Commented by mondodotto@gmail.com last updated on 13/Mar/18

$$\mathrm{please}\mathrm{recheck}$$

Commented by Tinkutara last updated on 13/Mar/18

Yes it is correct

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