If asinx = b sin [x + 2π/3] = c sin [ x + 4π /3] , prove that ab + bc + ca = 0.
Answers
Answer:
1/a + 1/b + 1/c = 0
Multiplying by abc on both sides we get
ab + bc + ca = 0 .... Hence proved.
Step-by-step explanation:
Let asinx = b sin [x + 2π/3] = c sin [ x + 4π /3] = k
therefore k/a = sin(x)
k/b = sin(x + + 2π/3)
k/c = sin (x + 4π /3)
Therefore k/a + k/b + k/c = sin(x) + sin(x + + 2π/3) + sin (x + 4π /3) = 0
Therefore k/a + k/b + k/c = 0
Therefore 1/a + 1/b + 1/c = 0
Multiplying by abc on both sides we get
ab + bc + ca = 0 .... Hence proved.
Answer:
1/a + 1/b + 1/c = 0
Multiplying by abc on both sides we get
ab + bc + ca = 0 .... Hence proved.
Step-by-step explanation:
Let asinx = b sin [x + 2π/3] = c sin [ x + 4π /3] = k
therefore k/a = sin(x)
k/b = sin(x + + 2π/3)
k/c = sin (x + 4π /3)
Therefore k/a + k/b + k/c = sin(x) + sin(x + + 2π/3) + sin (x + 4π /3) = 0
Therefore k/a + k/b + k/c = 0
Therefore 1/a + 1/b + 1/c = 0
Multiplying by abc on both sides we get
ab + bc + ca = 0 .... Hence proved.
Step-by-step explanation: