If \( \tan x=n \). tany, \( n \in R^{+} \), then maximum value of \( \sec ^{2}(x-y) \) is equal ...
If \( \tan x=n \). tany, \( n \in R^{+} \), then maximum value of \( \sec ^{2}(x-y) \) is equal to:
(A) \( \frac{(n+1)^{2}}{2 n} \)
(B) \( \frac{(n+1)^{2}}{n} \)
\( \mathrm{P} \)
(C) \( \frac{(n+1)^{2}}{2} \)
(D) \( \frac{(n+1)^{2}}{4 n} \)
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