f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π
f(A) = cosA = 1/(5) sin2B = 3/5 tanC= 1/3 cotD= 1/2
সমাধান কর: f(π/2-θ)f(π/2-3θ) =f(π/3) ;0<0<2π