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

If Sin θ = 1 2 , Then Find the Value of θ.

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प्रश्न

If sin θ = `1/2`, then find the value of θ. 

बेरीज
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उत्तर

sin θ = `1/2`

`sin 30^circ = 1/2`  ................ [using trignometric table]

∴ θ = 30°

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2018-2019 (July) Set 1

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संबंधित प्रश्‍न

Prove the following identities:

`(i) cos4^4 A – cos^2 A = sin^4 A – sin^2 A`

`(ii) cot^4 A – 1 = cosec^4 A – 2cosec^2 A`

`(iii) sin^6 A + cos^6 A = 1 – 3sin^2 A cos^2 A.`


Prove the following trigonometric identities.

`1/(sec A - 1) + 1/(sec A + 1) = 2 cosec A cot A`


Prove the following trigonometric identities.

`(1/(sec^2 theta - cos theta) + 1/(cosec^2 theta - sin^2 theta)) sin^2 theta cos^2 theta = (1 - sin^2 theta cos^2 theta)/(2 + sin^2 theta + cos^2 theta)`


Prove the following trigonometric identities.

`(1 + cot A + tan A)(sin A - cos A) = sec A/(cosec^2 A) - (cosec A)/sec^2 A = sin A tan A - cos A cot A`


Prove the following identities:

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


Prove the following identities:

`1/(secA + tanA) = secA - tanA`


Prove the following identities:

`(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(2sin^2A - 1)`


If x = r sin A cos B, y = r sin A sin B and z = r cos A, then prove that : x2 + y2 + z2 = r2


Prove that:

`cot^2A/(cosecA - 1) - 1 = cosecA`


`(1 + cot^2 theta ) sin^2 theta =1`


`If sin theta = cos( theta - 45° ),where   theta   " is   acute, find the value of "theta` .


Prove the following identity :

`1/(tanA + cotA) = sinAcosA`


Prove the following identity : 

`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`


Prove that: (1+cot A - cosecA)(1 + tan A+ secA) =2. 


Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.


Prove the following identities: cot θ - tan θ = `(2 cos^2 θ - 1)/(sin θ cos θ)`.


Prove that identity:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`


Complete the following activity to prove:

cotθ + tanθ = cosecθ × secθ

Activity: L.H.S. = cotθ + tanθ

= `cosθ/sinθ + square/cosθ`

= `(square + sin^2theta)/(sinθ xx cosθ)`

= `1/(sinθ xx  cosθ)` ....... ∵ `square`

= `1/sinθ xx 1/cosθ`

= `square xx secθ`

∴ L.H.S. = R.H.S.


(1 – cos2 A) is equal to ______.


`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.


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