Question Number 175544 by cortano1 last updated on 02/Sep/22
$$\:\mathrm{tan}\:^{\mathrm{6}} \left(\mathrm{10}°\right)+\mathrm{tan}\:^{\mathrm{6}} \left(\mathrm{50}°\right)+\mathrm{tan}\:^{\mathrm{6}} \left(\mathrm{70}°\right)=? \\ $$
Commented by Frix last updated on 29/Jan/23
$$\mathrm{433} \\ $$
Answered by a.lgnaoui last updated on 03/Sep/22
![posons t=tan(5°) tan 10^° =tan(2t) tan50^° =tan(45^° +t) tan70^° =tan(45^° +25^° ) t′=tan(25°) (((2t)/(1−t^2 )))^6 +(((1+t)/(1−t)))^6 +(((1−t′)/(1+t′)))^6 =((1/(1−t)))^6 [2^6 t^6 ((1/(1+t)))^6 +(1+t)^6 ]+(((1−t′)/(1+t′)))^6 =(1/((1−t)^6 ))[((64t^6 )/((1+t)^6 ))+(1+t)^6 ] t=0,087 t^′ =tan(25°)=(√2)−1=0,414 =(1/((0,913)^6 ))[((64×(0,087)^6 )/((1,087)^6 ))+(1,087)^6 ]+(((0,586)^6 )/8)=2,847+0,05=2,897 Resultat: tan 10 +tan 50 +tan 70=2,897](Q175577.png)
$${posons}\:\:\:{t}=\mathrm{tan}\left(\mathrm{5}°\right)\:\:\:\mathrm{tan}\:\mathrm{10}^{°} =\mathrm{tan}\left(\mathrm{2}{t}\right)\:\:\:\:\mathrm{tan50}^{°} =\mathrm{tan}\left(\mathrm{45}^{°} +{t}\right)\:\:\:\mathrm{tan70}^{°} =\mathrm{tan}\left(\mathrm{45}^{°} +\mathrm{25}^{°} \right)\:\:\:{t}'=\mathrm{tan}\left(\mathrm{25}°\right) \\ $$$$\left(\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }\right)^{\mathrm{6}} +\left(\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\right)^{\mathrm{6}} +\left(\frac{\mathrm{1}−{t}'}{\mathrm{1}+{t}'}\right)^{\mathrm{6}} \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}\right)^{\mathrm{6}} \left[\mathrm{2}^{\mathrm{6}} {t}^{\mathrm{6}} \left(\frac{\mathrm{1}}{\mathrm{1}+{t}}\right)^{\mathrm{6}} +\left(\mathrm{1}+{t}\right)^{\mathrm{6}} \right]+\left(\frac{\mathrm{1}−{t}'}{\mathrm{1}+{t}'}\right)^{\mathrm{6}} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}−{t}\right)^{\mathrm{6}} }\left[\frac{\mathrm{64}{t}^{\mathrm{6}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{6}} }+\left(\mathrm{1}+{t}\right)^{\mathrm{6}} \right] \\ $$$${t}=\mathrm{0},\mathrm{087}\:\:\:\:\:\:\:\:\:\:{t}^{'} =\mathrm{tan}\left(\mathrm{25}°\right)=\sqrt{\mathrm{2}}−\mathrm{1}=\mathrm{0},\mathrm{414} \\ $$$$ \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\left(\mathrm{0},\mathrm{913}\right)^{\mathrm{6}} }\left[\frac{\mathrm{64}×\left(\mathrm{0},\mathrm{087}\right)^{\mathrm{6}} }{\left(\mathrm{1},\mathrm{087}\right)^{\mathrm{6}} }+\left(\mathrm{1},\mathrm{087}\right)^{\mathrm{6}} \right]+\frac{\left(\mathrm{0},\mathrm{586}\right)^{\mathrm{6}} }{\mathrm{8}}=\mathrm{2},\mathrm{847}+\mathrm{0},\mathrm{05}=\mathrm{2},\mathrm{897} \\ $$$${Resultat}: \\ $$$$\mathrm{tan}\:\mathrm{10}\:+\mathrm{tan}\:\mathrm{50}\:+\mathrm{tan}\:\mathrm{70}=\mathrm{2},\mathrm{897} \\ $$$$\:\: \\ $$