Tan[sqrt(3) radians] = tan(1.7321) = -6.1475
Blank = pi/3 radians.
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tan θ = √3. tan-1(√3) = π/3 radians or 4π/3 radians, in degrees it would be 60° or 240° tan θ = sin θ / cos θ So, tan (π/3) = sin (π/3) / cos (π/3) = tan (4π/3) = sin (4π/3) / cos (4π/3) cos (π/3) = 1/2 cos (4π/3) = -1/2 Therefore cos θ = 1/2 or -1/2 when tan θ = √3
Tan of pi/2 + k*pi radians, for integer k, is not defined since tan = sin/cos and the cosine of these angles is 0. Since divsiion by 0 is not defined, the tan ratio is not defined.
Tan[sqrt(3) radians] = tan(1.7321) = -6.1475
Blank = pi/3 radians.
tan(60°) = sqrt(3)
tan(pi/3) = tan (60 degrees) = 1.732 which is square root of 3
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square root of 3
tan θ = √3. tan-1(√3) = π/3 radians or 4π/3 radians, in degrees it would be 60° or 240° tan θ = sin θ / cos θ So, tan (π/3) = sin (π/3) / cos (π/3) = tan (4π/3) = sin (4π/3) / cos (4π/3) cos (π/3) = 1/2 cos (4π/3) = -1/2 Therefore cos θ = 1/2 or -1/2 when tan θ = √3
Tan of pi/2 + k*pi radians, for integer k, is not defined since tan = sin/cos and the cosine of these angles is 0. Since divsiion by 0 is not defined, the tan ratio is not defined.
Tan(5.2 degrees) = 0.0910 Tan(5.2 radians) = -1.8856
Yes. (Theta in radians, and then approximately, not exactly.)
The answer, assuming that 120 is degrees and not radian is -square root of 3.
tan(6 radians) = -0.2910