Question

Prove the identity: \(\displaystyle \frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A.\)

Hint:

When expressions contain \(1 \pm \sin A\) or \(1 \pm \cos A\), try rationalization using conjugates to simplify quickly.

Answer 1

Concept: To prove trigonometric identities:

• Start from the more complicated side (usually LHS).
• Rationalize or combine terms using algebra.
• Use identities like \( \sin^2 A + \cos^2 A = 1 \) and \( \sec A = \frac{1}{\cos A} \).

Step 1: Start with LHS.
\[ \text{LHS} = \frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A}. \]
Step 2: Rationalize the first term.
Multiply numerator and denominator by \( 1 - \sin A \): \[ \frac{\cos A}{1 + \sin A} \cdot \frac{1 - \sin A}{1 - \sin A} = \frac{\cos A(1 - \sin A)}{1 - \sin^2 A}. \]
Step 3: Use identity \( 1 - \sin^2 A = \cos^2 A \).
\[ = \frac{\cos A(1 - \sin A)}{\cos^2 A} = \frac{1 - \sin A}{\cos A}. \]
Step 4: Substitute back into LHS.
\[ \text{LHS} = \frac{1 - \sin A}{\cos A} + \frac{1 + \sin A}{\cos A}. \]
Step 5: Combine the fractions.
\[ \text{LHS} = \frac{(1 - \sin A) + (1 + \sin A)}{\cos A} = \frac{2}{\cos A}. \]
Step 6: Convert to secant.
\[ \frac{2}{\cos A} = 2\sec A. \] Conclusion:
\[ \text{LHS} = \text{RHS} = 2\sec A. \] Hence, the identity is proved.