Question

(secA-tanA)^2 (1+sinA) = 1-sinA​

Answer 1

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Answer 2

Answer:

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