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Choose the Correct Alternative Answer for the Following Question. Cosec 45° = ? - Geometry Mathematics 2

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प्रश्न

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

पर्याय

  • \[\frac{1}{2}\]

  • \[\sqrt{2}\] 

  • \[\frac{\sqrt{3}}{2}\] 

  • \[\frac{2}{\sqrt{3}}\] 

MCQ
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उत्तर

\[cosec45^\circ = \sqrt{2}\]

Hence, the correct answer is \[\sqrt{2}\] .

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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 1.2 | पृष्ठ १३८

संबंधित प्रश्‍न

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.


If tanθ = 1 then, find the value of

`(sinθ + cosθ)/(secθ + cosecθ)`


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:

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

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.
sin \[\theta\] cosec \[\theta\]= ?


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.
cot2θ – tan2θ = cosec2θ – sec2θ


Prove the following.

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

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


Show that: 

`sqrt((1-cos"A")/(1+cos"A"))=cos"ecA - cotA"`


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


ΔAMT∼ΔAHE, construct Δ AMT such that MA = 6.3 cm, ∠MAT=120°, AT = 4.9 cm and `"MA"/"HA"=7/5`, then construct ΔAHE.


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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