To prove that \(\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A\) we will start from the left-hand side and manipulate it step by step.
Step 1: Start with the left-hand side
\(LHS = \sqrt{\frac{1 + \sin A}{1 - \sin A}}\)
Step 2: Multiply the numerator and denominator by 1+sin A
\(LHS = \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1-\sin A)(1 + \sin A)}}\)
This simplifies to:
\(LHS = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}}\)
Step 3: Use the identity \(1 - \sin^2 A = cos ^2 A\)
\(LHS = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}}\)
Step 4: Simplify the square root
\(LHS = \frac{1 + \sin A}{\cos A}\)
Step 5: Split the fraction
\(LHS = \frac{1}{\cos A} + \frac{\sin A}{\cos A}\)
Step 6: Write in terms of secant and tangent
\(LHS = \sec A + \tan A\)
Conclusion
Thus, we have shown that:
\(\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A\)
This completes the proof.
