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Prove the following. tan⁡θsec⁡θ+1=sec⁡θ−1tan⁡θ - Geometry Mathematics 2

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प्रश्न

Prove the following.

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

योग
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उत्तर

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

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

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

\[ = \frac{\tan\theta\left( \sec\theta - 1 \right)}{\tan^2 \theta} \]       ...(1 + tan2θ = sec2θ)

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

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अध्याय 6: Trigonometry - Problem Set 6 [पृष्ठ १३८]

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बालभारती Mathematics 2 [English] Standard 10 Maharashtra State Board
अध्याय 6 Trigonometry
Problem Set 6 | Q 5.08 | पृष्ठ १३८
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