Question

Prove the following trigonometric identities.

`(1 - sin θ)/(1 + sin θ) = (sec θ - tan θ)^2`

Prove that:

`(1 - sin θ)/(1 + sin θ) = (sec θ - tan θ)^2`

Answer 1

We have to prove  `(1 - sin θ)/(1 + sin θ) = (sec θ - tan θ)^2`

We know that, sin² θ + cos² θ = 1

Multiplying both numerator and denominator by  (1 − sin θ), we have

`(1 - sin θ)/(1 + sin θ) = ((1 - sin θ)(1 -  sin θ))/((1 + sin θ)(1 - sin θ))`

`= (1 - sin θ)^2/(1 - sin^2 θ)`

`= ((1 - sin θ)/cos θ)^2`

`= (1/cos θ - sin θ/cos θ)^2`

`= (sec θ - tan θ)^2`