Question

Prove that (secA-tanA)^2 (1+sinA)=1-sinA

Answer 1

hope you have your answer

Answer 2

Answer:

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