Question

Prove that \( \dfrac{1 + \sec A}{\sec A} = \dfrac{\sin^2 A}{1 - \cos A} \).

Hint:

Use trigonometric identities like \( \sin^2A = 1 - \cos^2A \) and cancel common terms carefully to prove equalities.

Answer 1

Step 1: Start with LHS.
\[ \text{LHS} = \frac{1 + \sec A}{\sec A} = \frac{1}{\sec A} + 1 = \cos A + 1 \]
Step 2: Simplify the RHS.
\[ \text{RHS} = \frac{\sin^2 A}{1 - \cos A} \] Use the identity \( \sin^2 A = 1 - \cos^2 A \): \[ \text{RHS} = \frac{1 - \cos^2 A}{1 - \cos A} = \frac{(1 - \cos A)(1 + \cos A)}{1 - \cos A} = 1 + \cos A \]
Step 3: Compare both sides.
\[ \text{LHS} = \text{RHS} = 1 + \cos A \]
Step 4: Conclusion.
Hence, \[ \boxed{\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 - \cos A}} \]