Question

1 - sinA /1+ sin A=(secA - tan A)²​

Answer 1

$\huge\purple {\mathfrak{Bonjour Mate!}}$

1-sinA/1+sinA=(secA-tanA)²

L.H.S

1-sinA/1+sinA

Rationalising the denominator

1-sinA (1-SinA) /1+sinA(1-SinA)

(1-SinA)²/ 1² -(Sin²A)

(1-SinA)²/ 1 -(Sin²A)

(1-SinA)² /Cos² A)

[ 1 -Sin²A = cos²A]

(1-SinA/CosA)²

(1/CosA-SinA/CosA)²

(SecA-tanA)²

L.H S = R H.S

Answer 2

Answer:

hello mate..

Step-by-step explanation:

1-sinA/1+sinA=(secA-tanA)²

L.H.S

1-sinA/1+sinA

Rationalising the denominator

1-sinA (1-SinA) /1+sinA(1-SinA)

(1-SinA)²/ 1² -(Sin²A)

(1-SinA)²/ 1 -(Sin²A)(1-SinA)² /Cos² A

[ 1 -Sin²A = cos²A]

(1-SinA/CosA)²

(1/CosA-SinA/CosA)²

(SecA-tanA)²

L.H.S= R.H.S

hope you get your answer

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