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

If tanθ = 1 then, find the value of θθθθsinθ+cosθsecθ+cosecθ

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प्रश्न

If tanθ = 1 then, find the value of

`(sinθ + cosθ)/(secθ + cosecθ)`

बेरीज
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उत्तर

tanθ = 1      ...(Given)
We know that, tan45° = 1
∴ θ = 45º

Now,

sinθ = sin 45º = `1/sqrt2`

cosθ = cos 45º = `1/sqrt2`

secθ = sec 45º = `sqrt2`

cosecθ = cosec 45º = `sqrt2`

∴ `(sinθ + cosθ)/(secθ + cosecθ)`

⇒ `(1/sqrt2 + 1/sqrt2)/(sqrt2 + sqrt2)`

⇒ `(2/sqrt2)/(2sqrt2)`

⇒ `cancel2/sqrt2 × 1/(cancel2sqrt2)`

⇒ `1/2`

∴ `(sinθ + cosθ)/(secθ + cosecθ) = 1/2`

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

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बालभारती Geometry Mathematics 2 [English] Standard 10 Maharashtra State Board
पाठ 6 Trigonometry
Practice Set 6.1 | Q 5 | पृष्ठ १३१

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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:

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


Prove that:

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

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


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 \[\theta\]  = ?


Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ 


Prove the following:

sec6x – tan6x = 1 + 3sec2x × tan2x


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.


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


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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)]`

 = `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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