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महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

Choose the Correct Alternative: Sinθ × Cosecθ =? - Geometry Mathematics 2

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प्रश्न

Choose the correct alternative: 
sinθ × cosecθ =?

पर्याय

  • `sqrt(2)`

  • `1/2`

  • 0

  • 1

MCQ

उत्तर

sinθ × cosecθ = 1

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Application of Trigonometry
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2018-2019 (July) Set 1

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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.


If tanθ = 1 then, find the value of

`(sinθ + cosθ)/(secθ + cosecθ)`


Prove that:

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


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


Prove that:

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

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


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.
cosec 45° =?


Prove the following.

secθ (1 – sinθ) (secθ + tanθ) = 1


Prove the following.
cot2θ – tan2θ = cosec2θ – sec2θ


Prove the following.

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

Prove the following.

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

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"`


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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