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

Prove the following.cot2θ – tan2θ = cosec2θ – sec2θ

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प्रश्न

Prove the following.
cot2θ – tan2θ = cosec2θ – sec2θ

बेरीज
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उत्तर

L.H.S = \[\cot^2 \theta - \tan^2 \theta\]

[1 + tan2θ = sec2θ, 1 + cot2θ = coses2θ]

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

\[ = {cosec}^2 \theta - 1 - \sec^2 \theta + 1\]

\[ = {cosec}^2 \theta - \sec^2 \theta\]

= R.H.S

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

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बालभारती Geometry Mathematics 2 [English] Standard 10 Maharashtra State Board
पाठ 6 Trigonometry
Problem Set 6 | Q 5.04 | पृष्ठ १३८

संबंधित प्रश्‍न

If \[\sin\theta = \frac{7}{25}\], find the values of cosθ and tan​θ.


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Prove that:

\[\cot\theta + \tan\theta = cosec\theta \sec\theta\]

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sin \[\theta\] cosec \[\theta\]= ?


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(secθ + tanθ) (1 – sinθ) = cosθ


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

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\[\frac{\tan^3 \theta - 1}{\tan\theta - 1} = \sec^2 \theta + \tan\theta\]

Choose the correct alternative: 
sinθ × cosecθ =?


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

Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)

= `(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))`

= `square/square xx 1/(square  square)`  ......`[(∵ square + square = 1),(∴ square = 1 - square)]`

 = `square/(square  square)`

= 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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