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Prove the following.cot2θ – tan2θ = cosec2θ – sec2θ - Geometry Mathematics 2

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प्रश्न

Prove the following.
cot2θ – tan2θ = cosec2θ – sec2θ

योग

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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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 5.04 | पृष्ठ १३८

संबंधित प्रश्न

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


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Prove that:

cos2θ (1 + tan2θ)


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

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(secθ - cosθ)(cotθ + tanθ) = tanθ.secθ.


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

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Prove that:

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

Choose the correct alternative answer for the following question.

When we see at a higher level, from the horizontal line, angle formed is ........
 

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

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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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