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

Simplify : 2 Sin30 + 3 Tan45.

Advertisements
Advertisements

प्रश्न

Simplify : 2 sin30 + 3 tan45.

Advertisements

उत्तर

2.sin30 + 3.tan45
= 2 × `1/2` + 3 × 1
= 1 + 3
= 4
shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2018-2019 (March) Balbharati Model Question Paper Set 3

संबंधित प्रश्‍न

Evaluate without using trigonometric tables:

`cos^2 26^@ + cos 64^@ sin 26^@ + (tan 36^@)/(cot 54^@)`


Prove the following trigonometric identities

cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1


Prove the following trigonometric identities.

`(cot A + tan B)/(cot B + tan A) = cot A tan B`


Prove the following identities:

`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`


Prove that:

2 sin2 A + cos4 A = 1 + sin4 A 


Prove that secθ + tanθ =`(costheta)/(1-sintheta)`.


Prove the following identity : 

`(1 + tan^2θ)sinθcosθ = tanθ`


Without using trigonometric identity , show that :

`sec70^circ sin20^circ - cos20^circ cosec70^circ = 0`


Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.


If A = 60°, B = 30° verify that tan( A - B) = `(tan A - tan B)/(1 + tan A. tan B)`.


Prove that `tan A/(1 + tan^2 A)^2 + cot A/(1 + cot^2 A)^2 = sin A.cos A`


Prove the following identities.

`sqrt((1 + sin theta)/(1 - sin theta)) + sqrt((1 - sin theta)/(1 + sin theta))` = 2 sec θ


If `cos theta/(1 + sin theta) = 1/"a"`, then prove that `("a"^2 - 1)/("a"^2 + 1)` = sin θ


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)


If sin θ + sin2 θ = 1 show that: cos2 θ + cos4 θ = 1


cos 45° = ?


If `cos θ = 24/25`, then sin θ = ?


Prove that `1/("cosec"  θ - cot θ) = "cosec"  θ + cot θ`.


Prove that `(sec A)/(tan A + cot A) = sin A`.


If sinθ = `11/61`, then find the value of cosθ using the trigonometric identity.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×