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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

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 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.


If 5 secθ – 12 cosecθ = 0, find the values of secθ, cosθ, and sinθ.


Prove that:

cos2θ (1 + tan2θ)


Prove that:

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

Prove that:

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


Prove that: `1/"sec θ − tan θ" = "sec θ + tan θ"`


Prove that:

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

Choose the correct alternative answer for the following question.
sin \[\theta\] cosec \[\theta\]= ?


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cosec 45° =?


Choose the correct alternative answer for the following question.

1 + tan2 \[\theta\]  = ?


Prove the following.

secθ (1 – sinθ) (secθ + tanθ) = 1


Prove the following.
(secθ + tanθ) (1 – sinθ) = cosθ


Prove the following:

sec6x – tan6x = 1 + 3sec2x × tan2x


Prove the following.

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

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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