The general value of θ in the equation cosθ=1√2,tanθ=−1 is
A
2nπ±π6,n∈I
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B
2nπ±7π4,n∈I
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C
nπ+(−1)n.π3,n∈I
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D
nπ+(−1)n.π4,n∈I
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Solution
The correct option is B2nπ±7π4,n∈I Here θ lies in the 4th quadrant. ∴θ=2π−π4=7π4
Now considering the equation, cosθ=cos7π4 ⇒ General value of θ=2nπ±7π4,n∈I