Question

prove that (secA- tanA)^2=1+sinA/1-sinA​

Answer 1

Answer:

(secA- tanA)2

Step-by-step explanation:

answer in the pic

Answer 2

Answer:

$(secA { - tanA})^{2} \\ \\ = ( \frac{1}{cosA} - \frac{sinA}{cosA} ) ^{2} \\ \\ = \frac{( {1 - sinA})^{2} }{ {cos}^{2} A} \\ \\ = \frac{( {1 - sinA})^{2} }{1 - {sin}^{2} A} \\ \\ = \frac{1 - sinA}{1 + sinA}$