हिंदी

Choose the Correct Alternative Answer for the Following Question. Sin θ Cosec θ = ? (A) 1 (B) 0 (C) 1 2 (D) √ 2

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प्रश्न

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

विकल्प

  • 1

  • 0

  • \[\frac{1}{2}\] 

  • \[\sqrt{2}\] 

MCQ
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उत्तर

\[\sin\theta cosec\theta\]
\[ = \sin\theta \times \frac{1}{\sin\theta}\]
\[ = 1\]
Hence, the correct answer is 1 .

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

APPEARS IN

बालभारती Geometry Mathematics 2 [English] Standard 10 Maharashtra State Board
अध्याय 6 Trigonometry
Problem Set 6 | Q 1.1 | पृष्ठ १३८

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.


Prove that:

cos2θ (1 + tan2θ)


Prove that:

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

Prove that:

\[\sec^4 A\left( 1 - \sin^4 A \right) - 2 \tan^2 A = 1\]

Choose the correct alternative answer for the following question.

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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.
cot2θ – tan2θ = cosec2θ – sec2θ


Prove the following.

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


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

Choose the correct alternative: 
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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)]`

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= 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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