English
Maharashtra State BoardSSC (English Medium) 10th Standard

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

Advertisements
Advertisements

Question

If tanθ = 1 then, find the value of

`(sinθ + cosθ)/(secθ + cosecθ)`

Sum
Advertisements

Solution

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
  Is there an error in this question or solution?
Chapter 6: Trigonometry - Practice Set 6.1 [Page 131]

APPEARS IN

Balbharati Geometry Mathematics 2 [English] Standard 10 Maharashtra State Board
Chapter 6 Trigonometry
Practice Set 6.1 | Q 5 | Page 131

RELATED QUESTIONS

If \[\sin\theta = \frac{7}{25}\], find the values of cosθ and tan​θ.


If \[\tan \theta = \frac{3}{4}\], find the values of sec​θ and cos​θ


Prove that:

cos2θ (1 + tan2θ)


Prove that:

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


Prove that:

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

Prove that:

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

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.
(secθ + tanθ) (1 – sinθ) = cosθ


Prove the following.

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

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 =?


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×