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

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Question

Prove that:

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

Solution

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

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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: 6.06 | Page no. 131
Solution Prave That: 1 Sec θ − Tan θ = Sec θ + Tan θ Concept: Application of Trigonometry.
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