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Question Number 141423 by naka3546 last updated on 18/May/21

tan (x+y)= ((12)/5)  sin (x−y) = (3/5)  x+y , x−y  are  acute  angles .  tan x tan y =  ?

tan(x+y)=125sin(xy)=35x+y,xyareacuteangles.tanxtany=?

Answered by mr W last updated on 18/May/21

u=tan x, v=tan y  ((u+v)/(1−uv))=((12)/5) ⇒u+v=((12(1−uv))/5)  tan (x−y)=(3/4)  ((u−v)/(1+uv))=(3/4) ⇒u−v=((3(1+uv))/4)  u=(1/2)(((12(1−uv))/5)+((3(1+uv))/4))=((63−33uv)/(40))  u=(1/2)(((12(1−uv))/5)−((3(1+uv))/4))=((33−63uv)/(40))  uv=((63−33uv)/(40))×((33−63uv)/(40))  2079(uv)^2 −6658(uv)+2079=0  uv=((3329±2600)/(2079))=((77)/(27)) or  ((27)/(77))

u=tanx,v=tanyu+v1uv=125u+v=12(1uv)5tan(xy)=34uv1+uv=34uv=3(1+uv)4u=12(12(1uv)5+3(1+uv)4)=6333uv40u=12(12(1uv)53(1+uv)4)=3363uv40uv=6333uv40×3363uv402079(uv)26658(uv)+2079=0uv=3329±26002079=7727or2777

Commented by naka3546 last updated on 25/May/21

Thank you, sir.

Thankyou,sir.

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