If `aa n db` are positive quantities such that `a > b ,` the minimum value of `as e ctheta-bt a ntheta` is `2a b` (b) `sqrt(a^2-b^2)` (c) `a-b` (d) `sqrt(a^2+b^2)`

Options

A. 2ab
B. `sqrt(a^2-b^2)`
C. a-b
D. `sqrt(a^2+b^2)`

Answer

B

Explanation

Let `s=asectheta-btantheta`
`or btantheta+s=asectheta`
`or (a^2-b^2)tan^2theta-2bstantheta+(a^2-s^2)=0`
For `tantheta" to be real " 4b^2s^2-4(a^2-b^2)(a^2-s^2)ge0`
`or a^2s^2gea^2(a^2-b^2)`
`or sgesqrt(a^2-b^2)`
Therefore, the minimum value of s is `sqrt(A^2-B^2)`.