1. ## minimum positive value.

If $a,b$ are positive quantities and $a\geq b$.Then find Minimum positive value of $asec\theta-btan\theta$

2. $\displaystyle a \sec \theta \tan \theta - b \sec^{2} \theta = 0$

$\displaystyle \frac{a}{b} = \frac{\sec \theta}{\tan \theta}$

$\displaystyle \frac{a}{b} = \frac{1}{\sin \theta}$

$\displaystyle \frac{b}{a} = \sin \theta$

$\displaystyle \theta = \arcsin \frac {b}{a} + 2 \pi n, \ n = 0, \ \pm 1, \pm2, ...$

3. but ans is in terms of $a$ and $b$ i.e

$asec\theta-btan\theta \geq \sqrt{a^2-b^2}$

4. Hello, jacks!

but ans is in terms of $\,a$ and $\,b$

$a\sec\theta-b\tan\theta \:\geq\: \sqrt{a^2-b^2}$

Random Variable did the hard work for you . . .

$\text{We have: }\:f(\theta) \:=\:a\sec\theta - b\tan\theta$

$\text{He found that }\,f'(\theta) = 0\,\text{ when }\,\sin\theta \,=\,\dfrac{b}{a}$

. . $\text{Hence: }\;\sec\theta \,=\,\dfrac{a}{\sqrt{a^2-b^2}},\;\;\tan\theta \,=\,\dfrac{b}{\sqrt{a^2-b^2}}$

Substitute into the function:

. . $\displaystyle a\sec\theta - b\tan\theta \;\ge\;a\frac{a}{\sqrt{a^2-b^2}} - b\frac{b}{\sqrt{a^2-b^2}}$

. . . . . . . . . . . . . . $\displaystyle =\;\frac{a^2}{\sqrt{a^2-b^2}} - \frac{b^2}{\sqrt{a^2-b^2}}$

. . . . . . . . . . . . . . $\displaystyle =\;\frac{a^2-b^2}{\sqrt{a^2-b^2}} \;=\;\sqrt{a^2-b^2}$

$\text{Therefore: }\;a\sec\theta - b\tan\theta \;\ge \;\sqrt{a^2+b^2}$

5. Thanks Random variable and Sorobon.