हिंदी

If tanθ = 1 then, find the value of θθθθsinθ+cosθsecθ+cosecθ - Geometry Mathematics 2

Advertisements
Advertisements

प्रश्न

If tanθ = 1 then, find the value of

`(sinθ + cosθ)/(secθ + cosecθ)`

योग
Advertisements

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`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Trigonometry - Practice Set 6.1 [पृष्ठ १३१]

APPEARS IN

बालभारती Mathematics 2 [English] Standard 10 Maharashtra State Board
अध्याय 6 Trigonometry
Practice Set 6.1 | Q 5 | 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.


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.

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.
cot2θ – tan2θ = cosec2θ – sec2θ


Prove the following:

sec6x – tan6x = 1 + 3sec2x × tan2x


Prove the following.

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


Prove the following.

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

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.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×