Question

$\frac{\sec A-\tan A}{\sec A+\tan A}=1-2\sec A\tan A+2{\tan }^{2}A$.

• A. True
• B. False

Answer 1

The correct option is A True

$\frac{\text{sec}A-\tan A}{\text{sec}A+\tan A}=\frac{(\text{sec}A-\tan A)(\text{sec}A-\tan A)}{(\text{sec}A+\tan A)(\text{sec}A-\tan A)}$

$=\frac{(\text{sec}A-\tan A{)}^2}{{\text{sec}}^{2}A-{\tan }^{2}A}$

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

$={\text{sec}}^{2}A+{\tan }^{2}A-2\text{sec}A\tan A$

$=1+{\tan }^{2}A+{\tan }^{2}A-2\text{sec}A\tan A$

$=1-2\text{sec}A\tan A+2{\tan }^{2}A$

So the relation is $\text{True}$