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Question Number 191901 by mehdee42 last updated on 03/May/23

$$if0<\theta <{45}^{0}whichisbigger?$$ $$2tan\theta ortan2\theta$$

Answered by mr W last updated on 03/May/23

Commented bymr W last updated on 03/May/23

$$BC=\sqrt{{BD}^2+{DC}^2}>BD=AB$$ $$AB=\tan \theta$$ $$AC=\tan 2\theta =AB+BC>AB+AB=2AB=2\tan \theta$$ $$\Rightarrow \tan 2\theta >2\tan \theta$$

Commented bymehdee42 last updated on 03/May/23

$$verygood$$

Answered by Subhi last updated on 04/May/23

$$tan\left(2\theta \right)=\frac{2tan\left(\theta \right)}{1-{\left(tan\left(\theta \right)\right)}^2}$$ $$0<\theta <45$$ $${tan}^2\left(0\right)<{tan}^2\left(\theta \right)<1$$ $$0<{tan}^2\left(\theta \right)<1$$ $$1>1-{tan}^2\left(\theta \right)>0$$ $$\frac{2tan\left(\theta \right)}{1-{tan}^2\left(\theta \right)}>2tan\left(\theta \right)$$ $$$$

Commented bymehdee42 last updated on 03/May/23

$$verynice$$

Answered by ajfour last updated on 04/May/23

$$f\left(\theta \right)=\tan 2\theta -2\tan \theta$$ $$=2\tan \theta \left\{\frac{1-(1-\tan {}^2\theta )}{1-\tan {}^2\theta }\right\}$$ $$=\frac{2\tan {}^3\theta }{1-\tan {}^2\theta }$$ $$f\left(\theta \right)>0if0<\tan \theta <1$$ $$\Rightarrow \tan 2\theta >2\tan \theta$$

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