Question

Prove that minimum value of $a\sec x-b\tan x$ is $\sqrt{a^2-b^2}$ where a and b are + ive, a > b

Answer 1

If $u=a\sec x-b\tan x$
(u + b tan x)${}^2$ = a${}^2(1+ta{n}^{2}x)$
or tan${}^{2}x(a^2-b^2)-2butanx+a^2-u^2=0$
$\Delta \ge 0,-a^2(a^2-b^2-u^2)\ge 0$
or $a^2(u^2-a^2+b^2)\ge 0$
$\therefore u^2\ge (a^2-b^2)\Rightarrow u\ge \sqrt{(}{a}^2-b^2).$
$\therefore$ Min.value of u is $\sqrt{(}{a}^2-b^2).$