Question

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

Hint:

Rationalizing helps simplify complex roots in trigonometry quickly.

Answer 1

Step 1: Understanding the Concept:
To eliminate the square root, rationalize the expression inside by multiplying by the conjugate of the denominator.
Step 2: Detailed Explanation:
LHS = \( \sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} \)
Multiply numerator and denominator by \( (1 - \sin \theta) \) inside the root:
\[ \text{LHS} = \sqrt{\frac{(1 - \sin \theta)(1 - \sin \theta)}{(1 + \sin \theta)(1 - \sin \theta)}} \]
\[ \text{LHS} = \sqrt{\frac{(1 - \sin \theta)^2}{1 - \sin^2 \theta}} \]
Using the identity \( 1 - \sin^2 \theta = \cos^2 \theta \):
\[ \text{LHS} = \sqrt{\frac{(1 - \sin \theta)^2}{\cos^2 \theta}} = \frac{1 - \sin \theta}{\cos \theta} \]
Splitting the terms:
\[ \text{LHS} = \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} = \sec \theta - \tan \theta = \text{RHS} \]
Step 3: Final Answer:
LHS = RHS. Hence Proved.