Question

PROVE THE FOLLOWING IDENTITY :-

1-sinA/1+sinA=(secA-tanA)^2

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Answer 1

Answer:

Solution

Let us start from the LHS

1-sinA/1+sinA

On rationalising the denominator we get

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

= RHS

Hence proved