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Prove the Following. Tan θ S E C θ + 1 = S E C θ − 1 Tan θ - Geometry

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Prove the following.
\[\frac{\tan\theta}{sec\theta + 1} = \frac{sec\theta - 1}{\tan\theta}\]

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Solution

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

Concept: Application of Trigonometry
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APPEARS IN

Balbharati Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board
Chapter 6 Trigonometry
Problem Set 6 | Q 5.08 | Page 138
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