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Solve for the general value of θ: tan (θ) tan (120º -θ) tan (120º + θ) = 1/√3. |
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Answer» The value of (θ) in tan (θ) tan (120º -θ) tan (120º +θ) is nπ/3 +π /18, n∈Z For angles greater than 90 degrees we define the tangent ofθ by tan θ. |
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