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Q.

If a and b are positive quantities such that a > b , then minimum value of  asec⁡θ−btan⁡θ is

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a

2ab

b

a2-b2

c

a-b

d

a2+b2

answer is B.

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Detailed Solution

Let s=asec⁡θ−btan⁡θor  btan⁡θ+s=asec⁡θor  a2−b2tan2⁡θ−2bstan⁡θ+a2−s2=0 For tan⁡θ to be real 4b2s2−4a2−b2a2−s2≥0or  a2s2≥a2a2−b2or  s≥a2−b2Therefore, the minimum value of s is a2-b2
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