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Question

If tanx=ntany, nR+, then maximum value of sec2(xy) is 
  1. (n1)24n
  2. (n+1)24n
  3. (n1)24n2
  4. (n+1)24n2


Solution

The correct option is B (n+1)24n
Given:
tanx=ntany
We know that,
tan(xy)=tanxtany1+tanxtanytan(xy)=ntanytany1+ntan2ytan(xy)=(n1)tany1+ntan2ysec2(xy)1=(n1)2tan2y(1+ntan2y)2(1)

Now, using A.M.G.M. for 1,ntan2y, we get
1+ntan2y2ntany
Squaring both sides,
(1+ntan2y)24ntan2y(1+ntan2y)24ntan2y14ntan2y(1+ntan2y)2(n1)24n(n1)2tan2y(1+ntan2y)2
Using equation (1), we get
(n1)24nsec2(xy)11+(n1)24nsec2(xy)(n+1)24nsec2(xy)

Hence, the minimum value of sec2(xy) is (n+1)24n

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