Prove that:

(sec A - tan A)² (1 + sin A) = (1 - sin A)

To prove:

(sec A - tan A)² (1 + sin A) = (1 - sin A)

Solving L.H.S. of the above equation,

$\Rightarrow \text{(sec A - tan A)}^2 \text{(1 + sin A)} \\[1em] \Rightarrow \Big(\dfrac{1}{\text{cos A}} - \dfrac{\text{sin A}}{\text{cos A}}\Big)^2 \text{(1 + sin A)} \\[1em] \Rightarrow \Big(\dfrac{\text{1 - sin A}}{\text{cos A}}\Big)^2\text{(1 + sin A)} \\[1em] \Rightarrow \dfrac{(1 - \text{sin A})^2\text{(1 + sin A)}}{\text{cos}^2 A} \\[1em] \Rightarrow \dfrac{(1 - \text{sin A})^2\text{(1 + sin A)}}{1 - \text{sin}^2 A} \\[1em] \Rightarrow \dfrac{(1 - \text{sin A})^2\text{(1 + sin A)}}{\text{(1 - sin A)(1 + sin A)}} \\[1em] \Rightarrow \text{1 - sin A}.$

Since, L.H.S. = R.H.S.

Hence, proved that (sec A - tan A)² (1 + sin A) = (1 - sin A).