Question

prove that 1+sina/1-sina=(seca+tana)²​

Answer 1

$\implies \frac{tan \: A \: + sin \: A}{tan \: A - sin \: A} = \frac{sec \: A+ 1}{sec \: A - 1}$

$\implies \: \frac{sin \: A \: + sin \: A \: cos \: A}{sin \: A - sin \: A \: cos \: A}$

$\implies\frac{1 + cos \: A}{1 - cos \: A }$

$\implies \: \frac{cos \: A( \frac{1}{cos \: A + 1}) }{cos \: A( \frac{1}{cos \: A \: - 1}) }$

$\implies \frac{sec \: A \: + 1}{sec \: A \: - 1}$

Answer 2

Step-by-step explanation:

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