हिंदी

From the Figure Find the Value of Sinθ.

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प्रश्न

From the figure find the value of sinθ.

योग
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उत्तर

`sinθ = ("AB")/("AC")`

`sinθ = 3/5`

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2018-2019 (March) Balbharati Model Question Paper Set 1

संबंधित प्रश्न

Prove the following trigonometric identities.

`(1 + tan^2 A) + (1 + 1/tan^2 A) = 1/(sin^2 A - sin^4 A)`


Prove the following trigonometric identities.

`(cot A - cos A)/(cot A + cos A) = (cosec A - 1)/(cosec A + 1)`


Prove the following trigonometric identities

sec4 A(1 − sin4 A) − 2 tan2 A = 1


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


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


If `(cot theta ) = m and ( sec theta - cos theta) = n " prove that " (m^2 n)(2/3) - (mn^2)(2/3)=1`


If `cos theta = 2/3 , " write the value of" (4+4 tan^2 theta).`


Eliminate θ, if
x = 3 cosec θ + 4 cot θ
y = 4 cosec θ – 3 cot θ


\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to 


\[\frac{1 + \tan^2 A}{1 + \cot^2 A}\]is equal to


Prove the following identity :

secA(1 - sinA)(secA + tanA) = 1


Prove the following identity :

cosecθ(1 + cosθ)(cosecθ - cotθ) = 1


Prove the following identity :

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


Prove the following identity :

`(tanθ + sinθ)/(tanθ - sinθ) = (secθ + 1)/(secθ - 1)`


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


Prove the following identities.

sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1


Prove that `[(1 + sin theta - cos theta)/(1 + sin theta + cos theta)]^2 = (1 - cos theta)/(1 + cos theta)`


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 `(sin θ + tan θ)/(cos θ) = tan θ (1 + sec θ)`.


Find the value of sin2θ  + cos2θ

Solution:

In Δ ABC, ∠ABC = 90°, ∠C = θ°

AB2 + BC2 = `square`   .....(Pythagoras theorem)

Divide both sides by AC2

`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`

∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`

But `"AB"/"AC" = square and "BC"/"AC" = square`

∴ `sin^2 theta  + cos^2 theta = square` 


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