Question

Prove that \(\sqrt{\frac{1 - \sin A}{1 + \sin A}} = \frac{1}{\sec A + \tan A}\).

Hint:

Rationalization is the key to solving trigonometric identities involving square roots.
Multiplying the numerator and denominator by the conjugate of the denominator helps eliminate the root easily!

Answer 1

Correct Answer: 4

Step 1: Understanding the Question:
This is a trigonometric identity proof.
We are required to establish the equality between the Left-Hand Side (LHS), which is inside a square root, and the Right-Hand Side (RHS).
We will simplify LHS by rationalizing the denominator, and then simplify RHS using basic reciprocal relations.

Step 2: Key Formula or Approach:
We will use standard trigonometric identities:
1. \(\sin^2 A + \cos^2 A = 1 \implies 1 - \sin^2 A = \cos^2 A\)
2. \(\sec^2 A - \tan^2 A = 1 \implies (\sec A + \tan A)(\sec A - \tan A) = 1\)
3. Rationalization of fractions inside square roots.

Step 3: Detailed Explanation:
1. Simplify the Left-Hand Side (LHS): \[ \text{LHS} = \sqrt{\frac{1 - \sin A}{1 + \sin A}} \] To remove the square root, multiply both the numerator and denominator inside the root by the conjugate of the denominator, which is \((1 - \sin A)\): \[ \text{LHS} = \sqrt{\frac{(1 - \sin A)(1 - \sin A)}{(1 + \sin A)(1 - \sin A)}} \] \[ \text{LHS} = \sqrt{\frac{(1 - \sin A)^2}{1 - \sin^2 A}} \] Using the identity \(1 - \sin^2 A = \cos^2 A\): \[ \text{LHS} = \sqrt{\frac{(1 - \sin A)^2}{\cos^2 A}} \] Taking the square root: \[ \text{LHS} = \frac{1 - \sin A}{\cos A} \] Split the fraction into two parts: \[ \text{LHS} = \frac{1}{\cos A} - \frac{\sin A}{\cos A} \] Using definitions \(\sec A = \frac{1}{\cos A}\) and \(\tan A = \frac{\sin A}{\cos A}\): \[ \text{LHS} = \sec A - \tan A \quad \text{--- (Equation 1)} \] 2. Simplify the Right-Hand Side (RHS): \[ \text{RHS} = \frac{1}{\sec A + \tan A} \] Multiply the numerator and denominator by the conjugate \((\sec A - \tan A)\): \[ \text{RHS} = \frac{\sec A - \tan A}{(\sec A + \tan A)(\sec A - \tan A)} \] Using the difference of squares in the denominator: \[ \text{RHS} = \frac{\sec A - \tan A}{\sec^2 A - \tan^2 A} \] We know the identity \(\sec^2 A - \tan^2 A = 1\). Substitute this value: \[ \text{RHS} = \frac{\sec A - \tan A}{1} \] \[ \text{RHS} = \sec A - \tan A \quad \text{--- (Equation 2)} \] 3. Compare Equation 1 and Equation 2: \[ \text{LHS} = \text{RHS} = \sec A - \tan A \]

Step 4: Final Answer:
Since the LHS simplifies to \(\sec A - \tan A\) and the RHS also simplifies to \(\sec A - \tan A\), the identity is successfully proven.