Question

(secA - Tan A)² (1+ sin A) = 1-sin A​

Answer 1

Step-by-step explanation:

= (1-sinA)/(1+sinA) × (1-sinA)/(1-sinA) By Rationalization

= (1-sinA)²/(1+sinA)(1-sinA)

= (1-sinA)/(1+sinA) = LHS

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Alternatively,

LHS

= (sec A - tan A)²; express all as sine and cosine

= (1/cos A - sin A/cos A)²

= [(1 - sin A)/cos A)]²

= (1 - sin A)²/cos² A

= (1 - sin A)²/(1 - sin² A)

= (1 - sin A)²/[(1 - sin A)(1 + sin A)

= (1 - sin A)/(1 + sin A)

= RHS

hope it's help you....

Answer 2

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