Question

1 + sin A / 1-sin A + 1- sin A/ 1+ sin A = sec ²A+tan²A​

Answer 1

L.H.S. = √{(1 - sina)/(1 + sina)}

= √[{(1 - sina)(1 - sina)}/{(1 + sina)(1 - sina)}]

= √{(1 - sina)²/(1 - sin²a)}

= √{(1 - sina)²/cos²a}

= √{(1 - sina)/cosa}²

= (1 - sina)/cosa

= 1/cosa - sina/cosa

= seca - tana = R.H.S.

Hence proved.

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