Question

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

Hint:

For expressions like \(\sqrt{\frac{1 \pm \sin A}{1 \mp \sin A}}\) or \(\sqrt{\frac{1 \pm \cos A}{1 \mp \cos A}}\), rationalizing the denominator is almost always the correct first step.

Answer 1

Step 1: Understanding the Concept:
Rationalization of the denominator within a square root often helps simplify trigonometric expressions.
Step 2: Key Formula or Approach:
Use the identity: \(\cos^2 A = 1 - \sin^2 A\).
Definitions: \(\sec A = \frac{1}{\cos A}\), \(\tan A = \frac{\sin A}{\cos A}\).
Step 3: Detailed Explanation:
Taking the Left Hand Side (LHS):
\[ LHS = \sqrt{\frac{1 + \sin A}{1 - \sin A}} \]
Multiply both the numerator and the denominator by \((1 + \sin A)\):
\[ = \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}} \]
\[ = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}} \]
Using the identity \(1 - \sin^2 A = \cos^2 A\):
\[ = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} \]
Taking the square root:
\[ = \frac{1 + \sin A}{\cos A} \]
Split the fraction:
\[ = \frac{1}{\cos A} + \frac{\sin A}{\cos A} \]
\[ = \sec A + \tan A \]
This matches the Right Hand Side (RHS).
Step 4: Final Answer:
LHS = RHS. Hence proved.