If a and b are positive numbers such that $a > b,$ then the minimum value of ${\rm{a\;sec}}\;\theta - b\tan \theta \left( {0 < \theta < \frac{\pi }{2}} \right)$ is

$\frac{1}{{\sqrt {{a^2} - {b^2}} }}$

$\frac{1}{{\sqrt {{a^2} + {b^2}} }}$

JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Analyzing +4 | -1

If a and b are positive numbers such that

a > b,

then the minimum value of

\mathrm{a\;sec}\;\theta - b\tan \theta \left( {0 < \theta < \frac{\pi }{2}} \right)

View answer and explanation

Correct option: D

\sqrt {{a^2} - {b^2}}

Option D (

\sqrt{{{{a^2} - {b^2}}}}

) is correct because it represents the minimum value of the expression, found using differentiation and trigonometric identities.