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Prove the following. 1 1 − sin θ + 1 1 + sin θ = 2 sec 2 θ - Geometry

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Question

Prove the following.

\[\frac{1}{1 - \sin\theta} + \frac{1}{1 + \sin\theta} = 2 \sec^2 \theta\]

Solution

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

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

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

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

\[ = 2 \sec^2 \theta\]

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APPEARS IN

 Balbharati Solution for Balbharati Class 10 Mathematics 2 Geometry (2018 to Current)
Chapter 6: Trigonometry
Problem set 6 | Q: 5.06 | Page no. 138
Solution Prove the following. 1 1 − sin θ + 1 1 + sin θ = 2 sec 2 θ Concept: Application of Trigonometry.
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