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

To Prove:
\[ \sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}} = \sec\theta - \tan\theta \]

Step 1: Start with the LHS
\[ \sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}} \]
Multiply numerator and denominator inside the root by \(1 - \sin\theta\):
\[ = \sqrt{\frac{(1 - \sin\theta)^2}{(1 + \sin\theta)(1 - \sin\theta)}} \]
Step 2: Use identity
\[ (1 + \sin\theta)(1 - \sin\theta) = 1 - \sin^2\theta = \cos^2\theta \]
Substitute:
\[ = \sqrt{\frac{(1 - \sin\theta)^2}{\cos^2\theta}} \]
Take square root:
\[ = \frac{1 - \sin\theta}{\cos\theta} \]

Step 3: Split the fraction
\[ = \frac{1}{\cos\theta} - \frac{\sin\theta}{\cos\theta} \] \[ = \sec\theta - \tan\theta \]

Final Answer:
\[ \boxed{\sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}} = \sec\theta - \tan\theta} \] Hence proved.