If a and b are positive quantities such that a > b , then minimum value of asecθ−btanθ 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=asecθ−btanθor btanθ+s=asecθor a2−b2tan2θ−2bstanθ+a2−s2=0 For tanθ to be real 4b2s2−4a2−b2a2−s2≥0or a2s2≥a2a2−b2or s≥a2−b2Therefore, the minimum value of s is a2-b2