Question

1+sin/1–sin = (sec + tan)²

plz give me solution.....​

Answer 1

Answer:

Step-by-step explanation:

1+sin/1-sin=(sec + tan)²    

LHS (multiplying it by conjugate)

1+sin(1+sin)/1-sin(1+sin)

(1+sin)²/{1-(sin)²}

(1+sin)²/1-sin²

(1+sin)²/cos²

{(1/cos²)+(sin²/cos²)}²

(sec + tan)²   RHS

THERFORE LHS=RHS

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Answer 2

$\frac{1 + sin}{1 - sin} \\ \\ = \frac{1 + sin}{1 - sin} \times \frac{1 + sin}{1 + sin} \\ \\ = \frac{( {1 + sin})^{2} }{1 - {sin}^{2} } \\ \\ = \frac{( {1 + sin})^{2} }{ {cos}^{2} } \\ \\ = ( \frac{1 + sin}{cos} ) ^{2} \\ \\ = ( {sec + tan})^{2}$