If sinθ=12,tanθ=1√3,∀n∈I, then most general values of θ are:
A
2nπ+π6,∀n∈I
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B
2nπ+π4,∀n∈I
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C
2nπ+π3,∀n∈I
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D
2nπ+π2,∀n∈I
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Solution
The correct option is A2nπ+π6,∀n∈I ∵sinθ=12=sinπ6 ⇒θ=π6,π−π6 and tanθ=1√3=tan(π6) ⇒θ=π6,π+π6 ∴ Common value of θ is π6 ∴ General value of θ is 2nπ+π6,∀n∈I