Question

sec A-tan A/sec A +tan A
= 1-2 sec A tan A + 2 tan’^2A​

Answer 1

$\frac{secA - tanA}{secA + tanA} \\ \\ = \frac{secA - tanA}{secA + tanA} \times \frac{secA - tanA}{secA - tanA} \\ \\ = \frac{( {secA - tanA})^{2} }{ {sec}^{2}A - {tan}^{2} A } \\ \\ = ( {secA - tanA})^{2} \\ \\ = {sec}^{2} A + {tan}^{2} A - 2secAtanA$