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

If sin θ = 11/61, find the values of cos θ using trigonometric identity.

Advertisements
Advertisements

प्रश्न

If sin θ = `11/61`, find the values of cos θ using trigonometric identity.

बेरीज
Advertisements

उत्तर

sin θ = `11/61`  ...[Given]

We have,

sin2θ + cos2θ = 1

⇒ cos2θ = 1 − sin2θ

⇒ cos2θ = `1 - (11/61)^2`

⇒ cos2θ = `1 - 121/3721`

⇒ cos2θ = `(3721 - 121)/3721`

⇒ cos2θ = `3600/3721`

⇒ cos θ = `sqrt((60/61)^2)`    ...[Taking the square root of both sides]

⇒ cos θ = `60/61`

Thus, the value of cos θ is `60/61`.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Trigonometry - Problem Set 6 [पृष्ठ १३८]

APPEARS IN

बालभारती Geometry Mathematics 2 [English] Standard 10 Maharashtra State Board
पाठ 6 Trigonometry
Problem Set 6 | Q 2 | पृष्ठ १३८

संबंधित प्रश्‍न

Prove the following trigonometric identities.

`(tan^2 A)/(1 + tan^2 A) + (cot^2 A)/(1 + cot^2 A) = 1`


If sin A + cos A = m and sec A + cosec A = n, show that : n (m2 – 1) = 2 m


If sin A + cos A = p and sec A + cosec A = q, then prove that : q(p2 – 1) = 2p.


Prove that:

(tan A + cot A) (cosec A – sin A) (sec A – cos A) = 1


If `( cosec theta + cot theta ) =m and ( cosec theta - cot theta ) = n, ` show that mn = 1.


If`( 2 sin theta + 3 cos theta) =2 , " prove that " (3 sin theta - 2 cos theta) = +- 3.`


If \[\cos A = \frac{7}{25}\]  find the value of tan A + cot A. 


If x = a cos θ and y = b sin θ, then b2x2 + a2y2 =


9 sec2 A − 9 tan2 A is equal to


The value of sin ( \[{45}^° + \theta) - \cos ( {45}^°- \theta)\] is equal to 


Prove the following identity :

`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`


Prove that  `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ = 1/(sin^2 θ. cos^2 θ) - 2`.


Prove that sin2 5° + sin2 10° .......... + sin2 85° + sin2 90° = `9 1/2`.


If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, then prove that x2 + y2 = 1


If sec θ = `25/7`, find the value of tan θ.

Solution:

1 + tan2 θ = sec2 θ

∴ 1 + tan2 θ = `(25/7)^square`

∴ tan2 θ = `625/49 - square`

= `(625 - 49)/49`

= `square/49`

∴ tan θ = `square/7` ........(by taking square roots)


Prove that `(cos^2θ)/(sinθ) + sin θ = "cosec"  θ`.


Prove that sec2θ – cos2θ = tan2θ + sin2θ.


The value of the expression [cosec(75° + θ) – sec(15° – θ) – tan(55° + θ) + cot(35° – θ)] is ______.


Simplify (1 + tan2θ)(1 – sinθ)(1 + sinθ)


If tan θ + sec θ = l, then prove that sec θ = `(l^2 + 1)/(2l)`.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×