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

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प्रश्न

Prove that:

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

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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Application of Trigonometry
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Trigonometry - Practice Set 6.1 [पृष्ठ १३१]

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बालभारती Geometry (Mathematics 2) [English] 10 Standard SSC Maharashtra State Board
अध्याय 6 Trigonometry
Practice Set 6.1 | Q 6.03 | पृष्ठ १३१
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