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महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

Prave That: √ 1 − Sin θ 1 + Sin θ = Sec θ − Tan θ - Geometry Mathematics 2

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प्रश्न

Prove that:

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

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

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

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

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

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पाठ 6: Trigonometry - Practice Set 6.1 [पृष्ठ १३१]

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बालभारती Mathematics 2 [English] Standard 10 Maharashtra State Board
पाठ 6 Trigonometry
Practice Set 6.1 | Q 6.03 | पृष्ठ १३१
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