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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 \[\cot\theta = \frac{40}{9}\], find the values of cosecθ and sinθ.


If tanθ = 1 then, find the value of

`(sinθ + cosθ)/(secθ + cosecθ)`


Prove that:

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

Prove that:

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

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


Prove that:

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

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


Choose the correct alternative answer for the following question.
cosec 45° =?


Prove the following.

secθ (1 – sinθ) (secθ + tanθ) = 1


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


Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ 


Prove the following.

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

Prove the following.

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


If sinθ = `8/17`, where θ is an acute angle, find the value of cos θ by using identities.


In ΔPQR, ∠P = 30°, ∠Q = 60°, ∠R = 90° and PQ = 12 cm, then find PR and QR.


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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