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

Given that ((2tan15^0 )/(1+tan^2 15^0 ))=sin30^0 , find tan15^0 leaving the answer in surd form (radicals)

$$\mathrm{Given}\:\mathrm{that}\:\frac{\mathrm{2tan15}^{\mathrm{0}} }{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{15}^{\mathrm{0}} }=\mathrm{sin30}^{\mathrm{0}} ,\:\mathrm{find}\:\mathrm{tan15}^{\mathrm{0}} \\ $$$$\mathrm{leaving}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{surd}\:\mathrm{form}\:\left(\mathrm{radicals}\right) \\ $$

Answered by BaliramKumar last updated on 10/Sep/22

let tan15° = x ((2x)/(1+x^2 )) = (1/2) 1+x^2 = 4x x^2 − 4x + 1 = 0 x = ((4 ± (√(16−4)))/2) = ((4 ± 2(√3))/2) = 2 ± (√3) x = 2 + (√3) = 3.732_(invalid) x = 2 − (√3) = 0.2679 tan0°<tan15°<tan45° 0<tan15°<1 0<0.2679<1 ∴ tan15° = 2 − (√3) = 0.2679

$${let}\:\:\:\:\:\:{tan}\mathrm{15}°\:=\:{x} \\ $$$$\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{1}+{x}^{\mathrm{2}} \:=\:\mathrm{4}{x} \\ $$$${x}^{\mathrm{2}} \:−\:\mathrm{4}{x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$${x}\:=\:\frac{\mathrm{4}\:\pm\:\sqrt{\mathrm{16}−\mathrm{4}}}{\mathrm{2}}\:=\:\frac{\mathrm{4}\:\pm\:\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\:\mathrm{2}\:\pm\:\sqrt{\mathrm{3}} \\ $$$$\underset{\mathrm{invalid}} {\underbrace{{x}\:=\:\mathrm{2}\:+\:\sqrt{\mathrm{3}}\:=\:\mathrm{3}.\mathrm{732}}}\:\:\:\:\:\:\:\:\:\:\:{x}\:=\:\mathrm{2}\:−\:\sqrt{\mathrm{3}}\:=\:\mathrm{0}.\mathrm{2679} \\ $$$${tan}\mathrm{0}°<{tan}\mathrm{15}°<{tan}\mathrm{45}° \\ $$$$\mathrm{0}<{tan}\mathrm{15}°<\mathrm{1} \\ $$$$\mathrm{0}<\mathrm{0}.\mathrm{2679}<\mathrm{1} \\ $$$$\therefore\:{tan}\mathrm{15}°\:=\:\mathrm{2}\:−\:\sqrt{\mathrm{3}}\:=\:\mathrm{0}.\mathrm{2679} \\ $$

Commented by pete last updated on 10/Sep/22

thank you sir

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

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