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

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

Answer 1

Answer: (D)

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

Answer 2

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