\tan \\theta \tan ( 120^\circ - \\theta ) \tan ( 120^\circ+ \\theta) = { 1 } /{ \sqrt{ 3 } } , \\theta=

Correct Answer is : nπ3+π18,n∈Z

{ { n\\pi } } /{ 3 } + { \\pi } /{ 18 } , { n } \in { Z }

{ { n\\pi } } /{ 3 } + { \\pi } /{ 12 } , { n } \in { Z }

{ { n\\pi } } /{12 } + { \\pi } /{ 12 } , { n } \in { Z }

{ { n\\pi } } /{ 3 } + { \\pi } /{6 } , { n } \in { Z }

AP and TS EAMCET Engineering Model Paper 8

Question : 19 of 160

Marks: +1, -0

tan‌θ‌tan(120°−θ)‌tan(120°+θ)=

√3

,θ=

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