Q. 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) $\sqrt{a^{2} - b^{2}}$
Explanation
(d)
Let $y = a$ sec$\theta - b$tan$\theta$
$\Rightarrow \ \ \ \ \frac{dy}{d\theta} = a\ sec\theta tan\theta - bsec^{2}\theta$
Put $\frac{dy}{d\theta} = 0 \Rightarrow sec\theta(a\ tan\theta - bsec\theta) = 0$
$\Rightarrow sin\theta = \frac{b}{a}\ \ \ \ \ \ \$($\because sec\theta \neq 0$)
Now, $\frac{d^{2}y}{d\theta^{2}} > 0,at\ sin\theta = \frac{b}{a}$
$\therefore$minimum value is
$y = a\frac{a}{\sqrt{a^{2} - b^{2}}} - b\frac{b}{\sqrt{a^{2} - b^{2}}}$
=$\sqrt{a^{2} - b^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.