Maharashtra State Board course SSC (English Medium) Class 10th Board Exam
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Prave That: √ 1 − Sin θ 1 + Sin θ = Sec θ − Tan θ - Geometry

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Question

Prove that:

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

Solution

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

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

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

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

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APPEARS IN

 Balbharati Solution for Balbharati Class 10 Mathematics 2 Geometry (2018 to Current)
Chapter 6: Trigonometry
Practice set 6.1 | Q: 6.03 | Page no. 131
Solution Prave That: √ 1 − Sin θ 1 + Sin θ = Sec θ − Tan θ Concept: Application of Trigonometry.
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