#### Question

Prove that:

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

#### Solution

\[\sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}}\]

\[ = \sqrt{\frac{1 - \sin\theta}{1 + \sin\theta} \times \frac{1 - \sin\theta}{1 - \sin\theta}}\]

\[ = \sqrt{\frac{\left( 1 - \sin\theta \right)^2}{1 - \sin^2 \theta}}\]

\[ = \sqrt{\frac{\left( 1 - \sin\theta \right)^2}{\cos^2 \theta}} \left( \cos^2 \theta + \sin^2 \theta = 1 \right)\]

\[= \frac{1 - \sin\theta}{\cos\theta}\]

\[ = \frac{1}{\cos\theta} - \frac{\sin\theta}{\cos\theta}\]

\[ = \sec\theta - \tan\theta\]

Is there an error in this question or solution?

Solution Prave That: √ 1 − Sin θ 1 + Sin θ = Sec θ − Tan θ Concept: Application of Trigonometry.