Temukan nilai dosa (a + b) jika tan a = 4/3 dan cot b = 5/12, 0 ^ derajat

Menjawab:

#sin (a + b) = 56/65 #

Diberikan, # tana = 4/3 dan cotb = 5/12 #

# rarrcota = 3/4 #

# rarrsina = 1 / csca = 1 / sqrt (1 + cot ^ 2a) = 1 / sqrt (1+ (3/4) ^ 2) = 4/5 #

# rarrcosa = sqrt (1-sin ^ 2a) = sqrt (1- (4/5) ^ 2) = 3/5 #

# rarrcotb = 5/12 #

# rarrsinb = 1 / cscb = 1 / sqrt (1 + cot ^ 2b) = 1 / sqrt (1+ (5/12) ^ 2) = 12/13 #

# rarrcosb = sqrt (1-sin ^ 2b) = sqrt (1- (12/13) ^ 2) = 5/13 #

Sekarang, #sin (a + b) = sina * cosb + cosa * sinb #

#=(4/5)(5/13)+(3/5)*(12/13)=56/65#

#sin (a + b) = 56/65 #

Sini, # 0 ^ circ <color (violet) (a) <90 ^ circ => I ^ (st) Kuadran => warna (biru) (Semua, fns.> 0. #

# 0 ^ circ <warna (violet) (b) <90 ^ circ => I ^ (st) Kuadran => warna (biru) (Semua, fns.> 0 #

Begitu, # 0 ^ circ <color (violet) (a + b) <180 ^ circ => I ^ (st) dan II ^ (nd) Kuadran #

# => warna (biru) (sin (a + b)> 0 #

Sekarang, # tana = 4/3 => seca = + sqrt (1 + tan ^ 2a) = sqrt (1 + 16/9) = 5/3 #

#:. warna (merah) (cosa) = 1 / detik = warna (merah) (3/5 #

# => warna (merah) (sina) = + sqrt (1-cos ^ 2a) = sqrt (1-9 / 25) = warna (merah) (4/5 #

Juga, # cotb = 5/12 => cscb = + sqrt (1 + cot ^ 2b) = sqrt (1 + 25/144) = 13/12 #

#:. warna (merah) (sinb) = 1 / cscb = warna (merah) (12/13 #

# => warna (merah) (cosb) = + sqrt (1-sin ^ 2b) = sqrt (1-144 / 169) = warna (merah) (5/13 #

Karenanya, #sin (a + b) = sinacosb + cosasinb #

# => sin (a + b) = 4 / 5xx5 / 13 + 3 / 5xx12 / 13 #

#sin (a + b) = 20/65 + 36/65 = 56/65 #