Answer: see proof below
Step-by-step explanation:
[tex]\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}[/tex]
Use the following Identities:
sec Ф = 1/cos Ф
cos² Ф + sin² Ф = 1
Proof LHS → RHS
[tex]\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}[/tex]
[tex]\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}[/tex]
[tex]\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}[/tex]
[tex]\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}[/tex]
[tex]\text{Identity:}\qquad \qquad \dfrac{\sin^2 A}{1-\cos A}[/tex]
[tex]\text{LHS = RHS:}\quad \dfrac{\sin^2 A}{1-\cos A}=\dfrac{\sin^2 A}{1-\cos A}\quad \checkmark[/tex]
