Question

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

• A. $ \frac{1}{\sqrt{{{a}^{2}}-{{b}^{2}}}} $
• B. $ \frac{1}{\sqrt{{{a}^{2}}+{{b}^{2}}}} $
• C. $ \sqrt{{{a}^{2}}+{{b}^{2}}} $
• D. $ \sqrt{{{a}^{2}}-{{b}^{2}}} $
• E. $ {{a}^{2}}-{{b}^{2}} $

Answer 1

Answer: (D)

Let $ y=a\sec \theta -b\tan \theta $
$ \Rightarrow $ $ \frac{dy}{d\theta }=a\sec \theta \tan \theta -b{{\sec }^{2}}\theta $
Put $ \frac{dy}{d\theta }=0\Rightarrow \sec \theta (a\tan \theta -b\sec \theta )=0 $
$ \Rightarrow $ $ \sin \theta =\frac{b}{a} $ $ (\because \sec \theta \ne 0) $ Now, $ \frac{{{d}^{2}}y}{d{{\theta }^{2}}}>0, $ at $ \sin \theta =\frac{b}{a} $
$ \therefore $ Minimum value is
$ y=a.\frac{a}{\sqrt{{{a}^{2}}-{{b}^{2}}}}-b.\frac{b}{\sqrt{{{a}^{2}}-{{b}^{2}}}} $
$=\sqrt{{{a}^{2}}-{{b}^{2}}} $