Question

If $\sec A+\tan A=m$ and $\sec A-\tan A=n$. Find the
value of $\sqrt{mn}$.

Answer 1

$\sec A+\tan A=m$ and $\sec A-\tan A=n$

$\sqrt{mn}=\sqrt{(\sec A+\tan A)(\sec A-\tan A)}$

using identity $(a-b)(a+b)=a^2-b^2$

$\sqrt{mn}=\sqrt{(\sec A{)}^2-(\tan A{)}^2}$

$\sqrt{mn}=\sqrt{{\sec }^{2}A-{\tan }^{2}A}$

as we knoe ${\sec }^{2}x={\tan }^{2}x+1$

hence, ${\sec }^{2}x-{\tan }^{2}x=1$

using above we have

$\sqrt{mn}=\sqrt{1}$

$\sqrt{mn}=1or-1$