If tanx =ntany, n e R+, then the maximum value of sec2(x-y) is equal to-

A \( \frac{\left ( n + 1 \right )^{2}}{\left ( 2 n \right.} \)

B \( \frac{\left ( n + 1 \right )^{2}}{n} \)

C \( \frac{\left ( n + 1 \right )^{2}}{2} \)

D \( \frac{\left ( n + 1 \right )^{2}}{\left ( 4 n \right.} \)

Solution

Correct Answer: Option D

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