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Prave That: Cot θ + Tan θ = C O S E C θ Sec θ - Geometry Mathematics 2

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प्रश्न

Prove that:

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

\[\cot\theta + \tan\theta\]

\[ = \frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta}\]

\[ = \frac{\sin^2 \theta + \cos^2 \theta}{\sin\theta\cos\theta}\]

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

\[ = \frac{1}{\sin\theta} \times \frac{1}{\cos\theta}\]

\[ = \text{ cosec } \theta\sec\theta\]

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पाठ 6: Trigonometry - Practice Set 6.1 [पृष्ठ १३१]

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बालभारती Mathematics 2 [English] Standard 10 Maharashtra State Board
पाठ 6 Trigonometry
Practice Set 6.1 | Q 6.05 | पृष्ठ १३१

संबंधित प्रश्‍न

If \[\tan \theta = \frac{3}{4}\], find the values of sec​θ and cos​θ


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


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


Prove that:

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

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° =?


Choose the correct alternative answer for the following question.

1 + tan2 \[\theta\]  = ?


Choose the correct alternative answer for the following question.

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

Prove the following.

secθ (1 – sinθ) (secθ + tanθ) = 1


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:

sec6x – tan6x = 1 + 3sec2x × tan2x


Prove the following.

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


Choose the correct alternative: 
sinθ × cosecθ =?


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


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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