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Prove the following. 1 1 − sin θ + 1 1 + sin θ = 2 sec 2 θ - Geometry Mathematics 2

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प्रश्न

Prove the following.

\[\frac{1}{1 - \sin\theta} + \frac{1}{1 + \sin\theta} = 2 \sec^2 \theta\]
योग

उत्तर

\[\frac{1}{1 - \sin\theta} + \frac{1}{1 + \sin\theta}\]

\[ = \frac{1 + \sin\theta + 1 - \sin\theta}{\left( 1 - \sin\theta \right)\left( 1 + \sin\theta \right)}\]

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

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

\[ = 2 \sec^2 \theta\]

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Application of Trigonometry
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Trigonometry - Problem Set 6 [पृष्ठ १३८]

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बालभारती Geometry (Mathematics 2) [English] 10 Standard SSC Maharashtra State Board
अध्याय 6 Trigonometry
Problem Set 6 | Q 5.06 | पृष्ठ १३८

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If \[\tan\theta + \frac{1}{\tan\theta} = 2\], then show that \[\tan^2 \theta + \frac{1}{\tan^2 \theta} = 2\]


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Prove that: (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ

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= `(1/square - cos θ) (square/square + square/square)` ......`[∵ sec θ = 1/square, cot θ = square/square and tan θ = square/square]`

= `((1 - square)/square) ((square + square)/(square  square))`

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= tan θ.sec θ

= R.H.S.

∴ L.H.S. = R.H.S.

∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ


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