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

Sin^2θ + sin^2(90 – θ) = ? A) 0 B) 1 C) 2 D) sqrt(2)

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प्रश्न

sin2θ + sin2(90 – θ) = ?

पर्याय

  • 0

  • 1

  • 2

  • `sqrt(2)`

MCQ
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उत्तर

1

Explanation:

(sin(90 – θ))2 = (cosθ)2 

sin2(90 – θ) = cos2θ   ...(1)

sin2θ + cos2θ = 1      

∴ sin2θ + sin2(90 – θ) = 1   ...From (1) 

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पाठ 6: Trigonometry - Exercise

संबंधित प्रश्‍न

Prove the following trigonometric identities:

`(\text{i})\text{ }\frac{\sin \theta }{1-\cos \theta }=\text{cosec}\theta+\cot \theta `


Prove the following identities, where the angles involved are acute angles for which the expressions are defined:

`(cosec  θ  – cot θ)^2 = (1-cos theta)/(1 + cos theta)`


Prove the following trigonometric identities.

`(cos theta)/(cosec theta + 1) + (cos theta)/(cosec theta - 1) = 2 tan theta`


Prove the following trigonometric identities.

(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)


If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that a2 + b2 = m2 + n2


Show that : `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec A cosec A`


Prove the following identities:

`((cosecA - cotA)^2 + 1)/(secA(cosecA - cotA)) = 2cotA`


`cot theta/((cosec  theta + 1) )+ ((cosec  theta +1 ))/ cot theta = 2 sec theta `


If sec θ + tan θ = x, then sec θ =


Prove the following identity : 

`sin^4A + cos^4A = 1 - 2sin^2Acos^2A`


If cosθ + sinθ = `sqrt2` cosθ, show that cosθ - sinθ = `sqrt2` sinθ.


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


Proved that cosec2(90° - θ) - tan2 θ = cos2(90° - θ)  +  cos2 θ.


Prove the following identities:

`(1 - tan^2 θ)/(cot^2 θ - 1) = tan^2 θ`.


Prove the following identities.

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


Prove the following identities.

sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1


If 3 sin θ = 4 cos θ, then sec θ = ?


Prove that `(cot A)/(1 - cot A) + (tan A)/(1 - tan A) = -1`.


Prove that sin4A – cos4A = 1 – 2 cos2A.


If tan θ – sin2θ = cos2θ, then show that `sin^2θ = 1/2`.


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