Question

$\frac{secA-tanA}{secA+tanA}=1-2secAtanA+2ta{n}^{2}A$

Answer 1

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