Question

prove that secA - tanA/secA + tanA = 1 - 2 secA tanA + 2 tan^2 A

Answer 1

Answer:

Hence proved

Step-by-step explanation:

Answer 2

Answer:

$\frac{secA - tanA}{secA + tanA} \\ \\ = \frac{secA - tanA}{secA + tanA} \times \frac{secA - tanA}{secA - tanA} \\ \\ = \frac{( {secA - tanA})^{2} }{ {sec}^{2}A - {tan}^{2} A } \\ \\ = ( {secA - tanA})^{2} \\ \\ = {sec}^{2} A + {tan}^{2} A - 2secAtanA \\ \\ = 1 + {tan}^{2} A+ {tan}^{2} A - 2secAtanA \\ \\ =1+2{tan}^{2}A-2secAtanA$