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Question
If tan θ + tan 2θ + tan 3θ = tan θ.tan 2θ. tan 3θ, then the general value of the θ is ______.
Options
nπ
`("n"pi)/6`
`"n"pi +- pi/4`
`("n"pi)/2`
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Solution
If tan θ + tan 2θ + tan 3θ = tan θ.tan 2θ. tan 3θ, then the general value of the θ is `("n"pi)/6`.
Explanation:
tan (A + B + C) = `(tan A + tan B - tanA. tan B. tan C)/(1 - tan A.tan B - tan B.tanC - tan C.tan A)`
Since, tan θ + tan 2θ + tan 3θ = tan θ.tan
2θ.tan 3θ,
we get, tan (θ + 2θ + 3θ) = θ
∴ tan 6θ = 0
∴ 6θ = nπ
θ = `(npi)/6`
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