tg(x-15°)ctg(x+15°)=1/3

Question

tg(x-15°)ctg(x+15°)=1/3

Answer 1

tg(x-15)*ctg(x+15)=1/3

tg(x-15)=(tg(x)-tg(15))/(1+tg(x)tg(15))

tg(x+15)=(tg(x)+tg(15))/(1-tg(x)tg(15))
ctg(x+15)=(1-tg(x)tg(15))/(tg(x)+tg(15))

tg(x-15)*ctg(x+15)=(tg(x)-tg(15))(1-tg(x)tg(15))/[(tg(x)+tg(15))(1+tg(x)tg(15))]=1/3

tg(30)=2tg(15)/(1-tg^2(15))=1/sqrt(3) пусть z=tg(15) тогда 
2z/(1-z^2)=1/sqrt(3) 
z1=2-sqrt(3) 
и 
z2=-2-sqrt(3)<0 z2 не лежит в первой четверти , значит угол не 15 градусов </span>

tg(x)=y
(y-(2-sqrt(3)))(1-y*(2-sqrt(3)))/[(y+2-sqrt(3))(1+y*(2-sqrt(3)))]=1/3
(y+2-sqrt(3))<>0 и (1+y*(2-sqrt(3)))<>0
3(y-(2-sqrt(3)))(1-y*(2-sqrt(3)))=[(y+2-sqrt(3))(1+y*(2-sqrt(3)))]
3(y-(2-sqrt(3)))(1-y*(2-sqrt(3)))-[(y+2-sqrt(3))(1+y*(2-sqrt(3)))]=0
4(sqrt(3)-2)(y-1)^2=0 => y=1

tg(x)=1 => x=pi/4+pi*k

Answer 2

tg(x-15°)*ctg(x+15°) = 1/3

tg(x-15* (П/180)) * ctg (x+15*п/180))= tg(12x-п)/12* сtg (12x+п)/12

tg(12x-п)/12*ctg(12x+п)/12=1/3

 

sin((12x-п:12)*сos((12x+п:12)

----------------------------------------=1/3

сos((12x-п:12)*sin((12x+п:12)

 

 x ∈ ∅;