Vyriešte 1 / (tan2x-tanx) -1 / (cot2x-cotx) = 1?

Vyriešte 1 / (tan2x-tanx) -1 / (cot2x-cotx) = 1?
Anonim

# 1 / (tan2x-Tanx) -1 / (cot2x-cotx) = 1 #

# => 1 / (tan2x-Tanx) -1 / (1 / (tan2x) -1 / Tanx) = 1 #

# => 1 / (tan2x-Tanx) + 1 / (1 / (Tanx) -1 / (tan2x)) = 1 #

# => 1 / (tan2x-Tanx) + (tanxtan2x) / (tan2x-Tanx) = 1 #

# => (1 + tanxtan2x) / (tan2x-Tanx) = 1 #

# => 1 / tan (2x-x) = 1 #

# => Tan (x) = 1 = tan (pi / 4) #

# => X = NI + pi / 4 #

odpoveď:

# X = NI + pi / 4 #

vysvetlenie:

# Tan2x-Tanx = (sin2x) / (cos2x) -sinx / cosx = (sin2xcosx-cos2xsinx) / (cos2xcosx) #

= #sin (2x-x) / (cos2xcosx) = sinx / (cos2xcosx) #

a # Cot2x-cotx = (cos2x) / (sin2x) -cosx / sinx = (sinxcos2x-cosxsin2x) / (sin2xsinx) #

= #sin (x-2x) / (sin2xsinx) = - sinx / (sin2xsinx) #

z toho dôvodu # 1 / (tan2x-Tanx) -1 / (cot2x-cotx) = 1 # možno písať ako

# (Cos2xcosx) / sinx + (sin2xsinx) / sinx = 1 #

alebo # (Cos2xcosx + sin2xsinx) / sinx = 1 #

alebo #cos (2x-x) / sinx = 1 #

alebo # Cosx / sinx = 1 # tj. # Cotx = 1 = lôžko (pi / 4) #

z toho dôvodu # X = NI + pi / 4 #