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Question

If tan x =n tany, nR+, then maximum value of sec2(xy)=___



A

(n+1)22n

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B

(n+1)2n

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C

(n+1)22

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D

(n+1)24n

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Solution

The correct option is D

(n+1)24n


tan x=n tan y
cos (x-y)=cosx cosy +sin x sin y
=cos x cos y (1+ tan x tan y)
=cosxcosy(1+ntan2y)sec2(xy)=sec2xsec2y(1+ntan2y)2=(1+tan2x)(1+tan2y)(1+ntan2y)2=(1+n2tan2y)(1+tan2y)(1+ntan2y)2=1+(n1)2 tan2y(1+n tan2y)2
Now, (1+ntan2y2)2ntan2y
tan2y(1+tan2y)214nsec2(xy)1+(n1)24n=(n+1)24n.


Mathematics

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