Question

If \( 0^\circ < A < 90^\circ \), then simplify \( \sqrt{\frac{1+\sin A}{1-\sin A}} \) “‘latex id="76r4p8"

• A. \( \sec A + \tan A \)
• B. \( \sin A + \tan A \)
• C. \( \sec A + \cos A \)
• D. \( \cot A + \tan A \)

Hint:

If you see \( \sqrt{\frac{1+\sin A}{1-\sin A}} \), immediately try converting denominator using \(1-\sin^2 A = \cos^2 A\). This is a standard JEE/board-level identity pattern.

Answer 1

The Correct Option is A

Concept: This type of expression is simplified using algebraic manipulation and trigonometric identities. The key identity used is: \[ 1 - \sin^2 A = \cos^2 A \] Also, rationalization using conjugates helps convert square-root expressions into standard trigonometric forms.

Step 1: Start with the given expression. \[ \sqrt{\frac{1+\sin A}{1-\sin A}} \]

Step 2: Eliminate the denominator difficulty using conjugate idea. Multiply numerator and denominator by \(1+\sin A\): \[ \sqrt{\frac{1+\sin A}{1-\sin A} \cdot \frac{1+\sin A}{1+\sin A}} \]

Step 3: Expand the expression carefully. Numerator becomes: \[ (1+\sin A)^2 \] Denominator becomes: \[ (1-\sin A)(1+\sin A) \] So expression becomes: \[ \sqrt{\frac{(1+\sin A)^2}{1-\sin^2 A}} \]

Step 4: Apply trigonometric identity. We know: \[ 1 - \sin^2 A = \cos^2 A \] So: \[ \sqrt{\frac{(1+\sin A)^2}{\cos^2 A}} \]

Step 5: Apply square root separately. \[ \frac{1+\sin A}{\cos A} \] (because \(0^\circ < A < 90^\circ\), all values are positive, so modulus is not required)

Step 6: Split into two parts. \[ \frac{1}{\cos A} + \frac{\sin A}{\cos A} \]

Step 7: Convert into standard trigonometric functions. \[ \sec A + \tan A \] Final Answer: \[ \boxed{\sec A + \tan A} \]