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

Riješite 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 (2 x-x) = 1 #

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

# => X = NPI + pi / 4 #

Odgovor:

# X = NPI + pi / 4 #

Obrazloženje:

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

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

i # Cot2x-Cotx = (cos2x) / (sin2x) -cosx / sinx = (sinxcos2x-cosxsin2x) / (sin2xsinx) #

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

Stoga # 1 / (tan2x-tanx) -1 / (cot2x-Cotx) = 1 # može biti napisan kao

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

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

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

ili # Cosx / sinx = 1 # tj # Cotx = 1 = ležaj (pi / 4) *

Stoga # X = NPI + pi / 4 #