Bagaimana saya boleh menyelesaikannya ??

Jawapan:

# (tan315-tan30) / (1 + tan315tan30) = - (2 + sqrt (3)) #

Penjelasan:

#rar (tan315-tan30) / (1 + tan315tan30) #

# = tan (315-30) #

# = tan285 #

# = tan (270 + 15) #

# = - cot15 #

# = - 1 / tan15 #

# = - 1 / tan (45-30) #

# = - 1 / ((tan45-tan30) / (1 + tan45tan30)) #

# = (tan30 + 1) / (tan30-1) #

# = (1 / sqrt3 + 1) / (1 / sqrt3-1) #

# = (1 + sqrt (3)) / (1-sqrt (3)) #

# = (1 + sqrt (3)) ^ 2 / (- 2) = - (2 + sqrt (3)) #

Jawapan:

# -2-sqrt (3) #

Penjelasan:

Kami tahu itu, #tan (A-B) = (tanA-tanB) / (1 + tanAtanB) #

Jadi, = (tan315 ^ 0-tan30 ^ 0) / (1 + tan315 ^ 0tan30 ^ 0) = tan (315 ^ 0-30 ^ 0) = tan285 ^ 0 = tan (360 ^ 0-75 ^ 0) 0 = -2-sqrt3 #

ATAU

# tan315 ^ 0 = tan (270 ^ 0 + 45 ^ 0) = - tan45 ^ 0 = -1andtan30 ^ 0 = 1 / sqrt3 #

jadi, # (tan315 ^ 0-tan30 ^ 0) / (1 + tan315 ^ 0tan30 ^ 0) = (- 1-1 / sqrt3) / (1-1 * 1 / sqrt3)

= ((sqrt (3) +1) / (sqrt (3) -1)) * ((sqrt (3) +1) / (sqrt (3) -1)) = - (3 + 2sqrt) +1) / (3-1) = - (4 + 2sqrt3) / 2 = -2-sqrt3 #