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Question Number 175961 by pete last updated on 10/Sep/22

$$\mathrm{Given}\mathrm{that}\frac{{2\tan 15}^0}{1+{\tan }^2{15}^0}={\sin 30}^0,\mathrm{find}{\tan 15}^0$$$$\mathrm{leaving}\mathrm{the}\mathrm{answer}\mathrm{in}\mathrm{surd}\mathrm{form}\left(\mathrm{radicals}\right)$$

Answered by BaliramKumar last updated on 10/Sep/22

$$lettan15°=x$$$$\frac{2x}{1+x^2}=\frac{1}{2}$$$$1+x^2=4x$$$$x^2-4x+1=0$$$$x=\frac{4\pm \sqrt{16-4}}{2}=\frac{4\pm 2\sqrt{3}}{2}=2\pm \sqrt{3}$$$$\underset{\mathrm{invalid}}{\underbrace{x=2+\sqrt{3}=3.732}}x=2-\sqrt{3}=0.2679$$$$tan0°<tan15°<tan45°$$$$0<tan15°<1$$$$0<0.2679<1$$$$\therefore tan15°=2-\sqrt{3}=0.2679$$

Commented by pete last updated on 10/Sep/22

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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